Yearly Archives: 2021

Question 1.
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Solution:
The object is kept between ƒ and C. So the image should be real, inverted and beyond C. To locate the sharp image, screen should be placed at the position of image.

So, the image is inverted and magnified. Thus in order to locate the sharp image, the screen should be kept 54 cm in front of concave mirror and image on the screen will be observed real, inverted and magnified. If the candle is moved closer to the mirror, the real image will move away from the mirror, hence screen has to be shifted away from the mirror to locate the sharp image.

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Solution:
A convex mirror always form virtual image, which is erect and small in size of an object kept in front of it. Focal length of convex mirror ƒ = + 15 cm Object distance u = – 12 cm Using mirror formula

Therefore, image is virtual, formed at 6.67 cm at the back of the mirror.

It shows the image is erect, small in size and virtual. When the needle is moved farther from mirror the image also move towards focus and decreasing in size. As u approaches v approaches focus but never beyond ƒ.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope have to be moved to focus on the needle again?
Solution:
We know the relation

Now if the water is replaced by other liquid, the apparent depth will change and microscope will have to be further moved to focus the image. With new liquid

Now the microscope will have to shift from its initial position to focus on the needle again which is at 7.67 cm depth. Shift distance = 9.4 – 7.67 = 1.73 cm.

Question 4.
Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass- air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface (figure c).

Solution:
(a) Applying Snell’s law for the refraction from air to glass. Refractive index of glass w.r.t. air

(b) Now Snell’s law for the refraction from air to water

(c) Now the light beam is incident at an angle 45° from water to glass

Question 5.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source)
Solution:
As shown in the figure all those light rays which are incident on the surface at angle of incidence more than critical angle, does total internal reflection and are reflected back in water only. All those light rays which are incident below critical angle emerges out of surface bending away from normal. All those light beams which are incident at critical angle grazes the surface of water.

Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Solution:
When the light beam is incident from air on to the glass prism, the angle of minimum deviation is 40°. Refractive index of glass w.r.t. air.


Now the prism is placed in water, new angle of minimum deviation can be calculated.

Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Solution:
Both faces should be of same radius of curvature

So, the radius of curvature should be 22 cm for each face of lens.

Question 8.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20 cm, and
(b) a concave lens of focal length 16 cm?
Solution:
(a) The convex lens is placed in the path of convergent beam.

The image 1 is formed by further converging beams at a distance of 7.5 cm from lens.
(b) A concave lens is placed in the path of convergent’ beam, the concave lens further diverge the light.

The image I is formed by diverged rays at 48 cm away from concave lens.

Question 9.
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Solution:
Object of size 3 cm is placed 14 cm in front of concave lens.


So, the image is virtual, erect, of the size 1.8 cm and is located 8.4 cm from the lens on the same side as object. As the object is moved away from the lens, the virtual image moves towards the focus of the lens but never beyond it. The image also reduces in size as shift towards focus.

Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Solution:
Equivalent focal length of the combination

Hence, system will behave as a diverging lens of focal length 60 cm.

Question 11.
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Solution:
(a) We want the final image at least distance of distinct vision. Let the object in front of objective is at distance υ₀.

Now we can get required position of object in point of objective.

(b) We want the final image at infinity. Let us again assume the object in front of objective at distance υ_0.

The object distance υ_e for the eyepiece should be equal to ƒ_e = 6.25 cm to obtain final image at ∞. So, image distance of objective lens

Question 12.
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Solution:
The image is formed at least distance of distinct vision for sharp focus. The separation between two lenses will be υ₀ + |υ_e|

Let us find first υ₀ the image distance for objective lens.

Also we can find object distance for eyepiece υ_e as we know

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Solution:

Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m and the radius of lunar orbit is 3.8 × 10⁸ sm.
Solution:
(a) ƒ_o = 15 m and ƒ_e = 1.0 cm angular magnification by the telescope normal adjustment

(b) The image of the moon by the objective at lens is formed on its focus only as the moon is nearly infinite distance as compared to focal length.

Distance of object i.e., Radius of lunar orbit, R_o = 3.8 × 10⁸ cm Distance of image for objective lens i.e., focal length of objective lens ƒ_o = 15 m Radius of image of moon by objective lens can be calculated.

Question 15.
Use the mirror equation to deduce that:
(a) An object placed between f and 2f of a concave mirror produces a real image beyond 2 f.
(b) A convex mirror always produces a virtual image independent of the location of the object.
(c) The virtual image produced by a convex mirror is always diminished in size and is located Between the focus and the pole.
(d) An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
Solution:
(a) We know for a concave mirror ƒ < 0 [negative] and u < 0 [negative]

(b) For a convex mirror, ƒ > 0, always positive and object distance u < 0, always negative.

So, whatever be the value of u, a convex mirror always forms a virtual image.
(c) In convex mirror focal length is positive hence ƒ > 0 and for an object distance from mirror with negative sign (u < 0)

hence the image is located between pole and focus of the mirror

So, the image is virtual and diminished.
(d) In concave mirror, ƒ < 0 for object placed between focus and pole of concave mirror

Question 16.
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Solution:
The shift in the image by the thick glass slab can be calculated. Here, shift only depend upon thickness of glass slab and refractive index of glass.
Shift = Real thickness – Apparent of thickness

The answer does not depend on the location of the slab.

Question 17.
(a) Figure shows a cross-section of’light pipe’ made of a glass fiber of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?

Solution:

(a) Let us first derive the condition for total internal reflection. Critical angle for the interface of medium 1 and medium 2.

Condition for total internal reflection from core to cladding

Now, for refraction at first surface air to core.

Thus all incident rays which makes angle of incidence between 0° and 60° will suffer total internal reflection in the optical fiber.
(b) When there is no outer covering critical angle from core to surface.

Thus all incident rays at first surface 0° to 90° will suffer total internal reflection inside core.

Question 18.
Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we’see’a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake.
(d) Does the apparent depth of a tank of water change if viewed obliquely? if so, does the apparent depth increase of decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. is this fact of some use to a diamond cutter?
Solution:
(a) In this situation when rays are convergent behind the mirror, both plane mirror and convex mirror can form real images of virtual objects.

(b) Here, the retina is working as a screen, where the rays are converging, but this screen is not at the position of formed virtual image, in fact the reflected divergent rays are converged by the eye lens at retina. Thus, there is no contradiction.

(c) An observer in denser medium will observe the fisherman taller than actual height, due to refraction from rare to denser medium.

(d) Apparent depth decreases if viewed obliquely as compared to when observed near normally.
(e) As hence, sin   refractive

index of diamond is much greater than that of ordinary glass, hence critical angle C for diamond is much smaller (24°) as compared to that of glass (42°).
A skilled diamond cutter thus can take the advantage of such large range of angle of incidence available for total internal reflection 24° to 90°. The diamond can be cut with so many faces, to ensure that light entering the diamond does multiple total internal reflections before coming out. This behavior produce brilliance i.e., sparkling effect in the diamond.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Solution:
Let the object be placed x m in front of lens the distance of image from the lens is (3 – x) m.

Condition for image to be obtained on the screen, i.e.m real image. 9 – 12ƒ > 0 or 9 > 12ƒ or f < 0.75 m. so, maximum focal length is  0.75 m.

Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Solution:
The image of the object can be located on the screen for two positions of convex lens such that u and v are exchanged.
The separation between two positions of the lens is x = 20 cm. It can be observed from figure.

Question 21.
(a) Determine the ‘effective focal length of the combination of two lenses in question 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangement. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Solution:
(a)
(i) Let a parallel beam of light incident first on convex lens, refraction at convex lens


The parallel beam of light appears to diverge from a point 216 cm from the center of the two lens system.
(ii) Now let a parallel beam of light incident first on concave lens.

The image I1 will act as real object for convex lens at 28 cm.

Thus the parallel incident beam appears to diverge from a point 420 -4 = 416 cm on the left of the center of the two lens system. Hence the answer depend upon which side of the lens system the parallel beam is made incident. Therefore the effective focal length is different in two situations.

(b) Now an object of 1.5 cm size is kept 40 cm in front of convex lens in the same system of lenses.

Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
Solution:
The beam should be incident at critical angle or more than critical angle, for total internal reflection at second surface of the prism.

Let us first find critical angle for air glass interface.

Question 23.
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(a) deviate a pencil of white light without much dispersion.
(b) disperse (and displace) a pencil of white light without much deviation.
Solution:
By using two identical shape prism of crown glass and flint glass kept with their base on opposite sides, we can observe deviation without dispersion or dispersion without deviation.
(a) Deviation without dispersion A white beam incident on crown glass will suffer deviation 6, and angular dispersion Δθ, both

Now the light beam again suffer deviation and dispersion by flint glass prism.


Negative sign shows that two prisms must be placed with their base on opposite sides. Net deviation in this condition

(b) Dispersion without deviation

Here again negative sign shows that two prisms must be placed with their base on opposite sides. Net angular dispersion
Δθ = Δθ₁ + Δθ₂
Δθ = A₁ (μ_v1 – μ_R1) + A₂ (μ_v2 – μ_R2)

Question 24.
For a normal eye, the far points at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Solution:
Here the least converging power of eye lens is given as 20 dioptres behind the cornea. If we can calculate the maximum converging power, then we can get the range of accommodation of a normal eye.
(a) To see objects at infinity, the eye uses its least converging power = 40 + 20 = 60 dioptres
∴ Approximate distance between the retina and the cornea eyelens

(b) To focus object at the near point on the retina, we have

dioptres Power of the eye lens = 64 – 40 = 24 dioptres Thus the range of accommodation of the eye lens is approximately 20 to 24 dioptres.

Question 25.
Does short-sightedness (myopia) or long¬> sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of r accommodation? If not, what might cause these defects of vision?
Solution:
A person with normal ability of accommodation may be myopic or hypermetropic due to defective eye structure, j- When the eye ball from front to back gets too elongated the myopic defect occur, similarly when the eye ball from front to back gets too shortened the hypermetropia defect occur.
When the eye ball has normal length ‘ but the eye lens loses partially its ability of accommodation, the defect is called “presbyopia” and is corrected in the same I manner as myopia or hypermetropia.

Question 26.
A myopic person has been using spectacles of power – 1.0 dioptre for distant vision. During old age, he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.
Solution:
Myopic person uses spectacles of power – 1.0 dioptre or concave lens of focal length ƒ =  = -100 cm in order to observe clearly object at infinity. Far point of the person can be calculated as

Similarly if the person uses spectacles of power + 2.0 dioptre then he must be using convex lens of power + 2.0 dioptre. Focal length and near point can be calculated as

Thus the person also has the defect of hypermetropia and has a near point 50 cm. So having both defects he needs different lenses for distant vision and to see closer objects.

Question 27.
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
Solution:
This defect is called astigmatism. It arises due to non spherical cornea. The eye lens is ideally spherical and has same curvature in different planes, but in an astigmatic eye due to non spherical cornea the curvature may be insufficient in different planes.
In the given situation the curvature in the vertical plane is enough, so vertical lines are visible distinctly. But the curvature is insufficient in the horizontal plane, hence horizontal lines appear blurred. The defect can be corrected by using a cylindrical lens with its axis along vertical. The parallel rays in the vertical plane will suffer no extra refraction but the parallel rays in the horizontal plane will be refracted largely and converges at the retina, according to the requirement to form the clear image of horizontal lines.

Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
Solution:
(a) At closest distance of the object the image is formed at least distance of distinct vision and eye is most strained.

At farthest distance of the object the image is formed at ∞ and eye is most relaxed.

(b) Maximum angular magnification

Question 29.
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Solution:
(a) For magnification by the magnifying lens.

(b) Angular magnification,

(c) No, the linear magnification by a lens and magnifying power (angular magnification) of magnifying glass have different values. The linear magnification is calculated using

image of object at eye lens ‘p’ to the angle subtended by object assumed to be at least distance at eye lens ‘a’.
The linear magnification and angular magnification in microscope have similar magnitude when image is at least distance of distinct vision i.e., 25 cm.

Question 30.
(a) At what distance should the lens be held from the figure in previous question in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
Solution:
(a) For maximum magnifying power the image should be at least distance of distinct 1 vision i.e., 25 cm.

(b) Linear magnification in the situation of maximum magnifying power.

(c) Maximum magnifying power in the same situation

So, it can be observed that in the situation when image is least distance of distinct vision the angular magnification and linear magnification have similar values.

Question 31.
What should be the distance between the object in previous question and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier?
Solution:
Now we want the area of square shaped virtual image as 6.25 mm². So, each side of image is I =  = 2.5 mm (Linear magnification) For the given magnifying lens of focal length 10 cm we can calculate the required position of object.

Thus the required virtual image is closer than normal near point. Thus the eye cannot observe the image distinctly.

Question 32.
Answer the following question:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Solution:
(a) In magnifying glass the object is placed closer than 25 cm, which produces image at 25 cm. This closer object has larger angular size than the same object at 25 cm. In this way although the angle subtended by virtual image and object is same at eye but angular magnification is achieved.

(b) On moving the eye backward away from lens the angular magnification decreases slightly, as both the angle subtended by the
image at eye ‘a’ and by the object at eye ‘α’ decreases. Although the decrease in angle subtended by object a is relatively smaller.

(c) If we decrease focal length, the lens has to be thick with smaller radius of curvature. In a thick lens both the spherical aberrations and chromatic aberrations become pronounced. Further, grinding for small focal length is not easy. Practically we can not get magnifying power more than 3 with a simple convex lens.

(d) Magnifying power of a compound microscope is given by

(e) If we place our eye too close to the eyepiece, we shall not collect much of the light and also reduce our field of view. When we position our eye slightly away and the area of the pupil of our eye is greater, our eye will collect all the light refracted by the objective, and a clear image is observed by the eye.

Question 33.
An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Solution:
Here we want the distance between given objective and eye lens for the required magnification of 30. Let the final image is formed at least distance of distinct vision for eyepiece.

Question 34.
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when:
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?
Solution:
(a) In normal adjustment magnifying power

(b) For the image at least distance of distinct vision

Question 35.
(a) For the telescope described in previous question 34 (a), what is the separation between the objective and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away. What is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
Solution:
(a) The separation between objective lens and the eyepiece can be calculated in both the conditions of most relaxed eye and most strained eye. Most relaxed eye L= ƒ₀ + ƒ_e= 140 + 5 = 145 cm
Most strained eye object distance ‘u_e‘ for eye lens

(b)

(c) Now we want to find the height of final image A”B” assuming it to be formed at 25 cm.

Question 36.
A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Solution:
Image formed by concave mirror acts as a virtual object for convex mirror. Here parallel rays coming from infinity will focus at 110 mm an axis away from concave mirror. Distance of virtual object for convex mirror

Hence image is formed at 315 mm from convex mirror.

Question 37.
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in figure. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Solution:
If the mirror deflect by 3.5°, the reflected light deflect by 7°, deflection of the spot d can be calculated.

Question 38.
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Solution:
Let us first consider the situation when there is no liquid between lens and plane mirror and the image is formed at 30 cm i.e., at the position of object. As the image is formed on the object position itself, the object must be placed at focus of Biconvex lens.

ƒ₀ = 30 cm Radius of curvature of convex lens can be calculated

Now a liquid is filled between lens and plane mirror and the image is formed at position of object at 45 cm. The image is formed on the position of object itself, the object must be placed at focus of equivalent lens of Biconvex of glass and Plano convex lens of liquid

NCERT Exercises

Question 1.
Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Solution:

(a) Capacitance of the parallel plate capacitor

(b) The magnitude of displacement current is equal to conduction current. ∴ I_d = I = 0.15 A
(c) Yes, Kirchhoff’s first law is very much applicable to each plate of capacitor as l_d = I. So current is continuous and constant across each plate.

Question 2.
A parallel plate capacitor as shown in figure made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^-1.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of 6 at a point 3.0 cm from the axis between the plates.

Solution:
(a) Capacity of capacitor C = 100 pF


(b) Yes, the conduction current in wires is always equal to displacement current within plates.
(c)

To find magnetic field B of a point 3 cm from the axis within plates, let us assume a loop of radius 3 cm with center on axis. Now modified Ampere’s Law.

Question 3.
What physical quantity is same for X-rays of wavelength 10^-10 m, red light of wavelength 6800 A and radio waves of wavelength 500 m?
Solution:
Though they all have different wavelengths and frequencies, but they have same speed i.e., speed of light 3 × 108 m s^-1 in vacuum.

Question 4.
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Solution:
In electromagnetic wave, the electric field vector E and magnetic field vector B show their variations perpendicular to the direction of propagation of wave as well as perpendicular to each other. As the electromagnetic wave is travelling along z-direction, hence E and B show their variation in x-y plane.

Question 5.
A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Solution:
Maximum wavelength in the band is for lowest frequency

Minimum wavelength in the band is for highest frequency

Question 6.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of electromagnetic wave produced by the oscillator?
Solution:
The frequency of electromagnetic wave produced will be same as the frequency of oscillation of charged particles i.e., 109 Hz.

Question 7.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?
Solution:

Question 8.
Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N C^-1 and that its frequency is u = 50.0 MHz. (a) Determine B₀, to, k, and X.
(b) Find expressions for E and B.
Solution:
(a) By relation, 

(b) Let the wave is propagating along A’-direction, electric field E is along y – direction and magnetic field B along z-direction.

Question 9.
The terminology of different parts of the electromagnetic spectrum is given in the NCERT text. Use the formula E = hυ and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Solution:
The photon of given energy are released during transition between energy levels in the atom and the emitted photon energy is equal to difference of energies of energy levels, among which the transition has taken place. For example photon energy for
λ = 1 m can be calculated

So, for emission of radio waves the energy difference between energy levels should be 12.425 × 10^-7 eV. Similarly, photon energy for other wavelengths can also be calculated

So, for emission of y -rays the energy difference among energy levels should be of the order of MeV, where for visible radiation it should be of the order of eV.

Question 10.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and amplitude 48 Vm^-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field, [c = 3 × 10⁸ m s^-1.]
Solution:
(a) Wavelength of electromagnetic wave is

(b) Amplitude of magnetic field, 

(c) Energy density as electric field,

Question 11.
Suppose that the electric field part of an electromagnetic wave in vacuum is £ = [(3.1 N/C)] cos[(1.8 rad/m) y + (5.4 x 10⁸ rad/s)f)] i.
(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency υ?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
Solution:
(a) Electromagnetic wave is propagating in -y direction.
(b) Propagation constant, k =  = 1.8

(c) angular frequency, ω = 2πυ = 5.4 × 10⁸

(d) Amplitude of magnetic field pat, 

(e) Expression for magnetic field pat of wave

Question 12.
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted is o tropically and neglect reflection.
Solution:
The bulb as a point source, radiate light in all direction uniformly and it is given that only 5% of power is converted to visible radiations.
(a) Let us assume a sphere of radius 1 m. Surface area A = 4πr² = 4π (l)² = 4 π m² Intensity on the sphere

(b) Let us assume a sphere of 10 m radius. Surface area A = 4π² = 4K (10)² = 400 π m² Intensity on the sphere

Question 13.
Use the formula λ_mT = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Solution:
A black body at very high temperature produce a continuous spectrum. Using Wien’s displacement law we can calculate the wavelength corresponding to maximum intensity of radiation emitted Required absolute temperature

Using above formula, the temperature of black body required for various wavelength is calculated.

To produce electromagnetic radiations of different wavelength, we need temperature ranges. To produce visible ’ radiation of X = 5 × 10^-7 m, we need to have source at temperature of 6000 K. A source at lower temperature will produce this wavelength but not with maximum intensity.

Question 14.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in Physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm [wavelength emitted by atomic ‘hydrogen in interstellar space].
(b) 1057 MHz [frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift].
(c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
(d) 5890 A – 5896 A [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in ⁵⁷Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy)].
Solution:
(a) 21 cm wavelength corresponds to radio waves, at nearly 1445 MHz.
(b) Electromagnetic wave of 1057 MHz also corresponds to radio waves.
(c) Using Wien’s Law the wavelength of the radiation mostly emitted at given temperature of heavenly bodies can be calculated

This wavelength corresponds to microwave region of electromagnetic waves.
(d) This wavelength corresponds to visible region (yellow) of electromagnetic waves.
(e) The frequency of radiation can be calculated

This frequency corresponds to X-ray region of electromagnetic waves.

Question 15.
Answer the following questions:
(a) Long distance radio broadcasts use short¬wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?
Solution:
(a) Electromagnetic waves in frequency range of shortwave band reflect from ionosphere where lower frequency radio waves i.e., medium waves are absorbed. So, short waves are suitable for long distance radio broadcast.
(b) Very high frequency (VHF) and ultra high frequency (UHF) electromagnetic waves used in T.V transmission after frequency modulation do not reflect from ionosphere and can only be transmitted straight from antenna. So, the range is highly limited. In order to send the transmission for long distances, satellite is used.
(c) Atmosphere absorb X-rays, while visible and radio waves can penetrate through it. Hence optical telescope can work on ground but X-ray astronomical telescopes only work above atmosphere, hence installed on the satellite orbiting around earth.
(d) The thin ozone layer absorbs harmful ultraviolet radiations, y-rays and cosmic radiations. All these high energy radiations can cause damage to the living cells.
Thus the thin ozone layer on top of the stratosphere is crucial for human survival.
(e) In the absence of atmosphere, the day temperature can rise to many fold and temperature at night will be much below 0°C. Although the average temperature will be reduced in the absence of green house effect.
(f) The thick clouds produced by global nuclear war will prevent solar radiation to reach down across the globe. This would cause severe ‘nuclear winter’.

NCERT Exercises

Question 1.
A 100 Q resistor is connected to a 220 V, 50 Hz a.c. supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Solution:
(a) Here virtual a.c. voltage is 220 V at a frequency of 50 Hz. So, rms value of current


(b) Power in complete cycle

Question 2.
(a) The peak voltage of an a.c. supply is 300 V. What is the rms voltage?
(b) The rms value of current in an a.c. circuit is 10 A. What is the peak current?
Solution:
(a) The peak value of a.c. supply is given 300 V.

Question 3.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of current in the circuit.
Solution:

Question 4.
A 60 μF capacitor is connected to a 110 V, 60 Hz a.c. supply. Determine the rms value of current in the circuit.
Solution:

Question 5.
In previous questions 3 and 4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Solution:
For question 3, Power in the circuit with pure inductor P = Eυlυ cos π/2 = 0. For question 4, Power in complete cycle P = Eυlυ cos (-π/2) = 0.

Question 6.
Obtain the resonant frequency a), of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω2. What is the Q-value of this circuit?
Solution:
Resonant angular frequency in series LCR circuit

Question 7.
A charged 30 pF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Solution:
Angular frequency of LC oscillations

Question 8.
Suppose the initial charge on the capacitor given in question 7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Solution:
Initial energy on capacitor

Any time total energy in the circuit is constant, hence energy later is 0.6 J.

Question 9.
A series LCR circuit with R = 20 Ω 2, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Solution:

Average power transferred to the circuit in one complete cycle at resonance

Question 10.
A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz), if its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?
Solution:
For tuning, the natural frequency i.e., the frequency of L-C oscillations should be equal to frequency of radio waves received by the antenna in the form of same frequency current in the L-C circuit. For tuning at 800 kHz, required capacitance

So, the variable capacitor should have a frequency range between 87.9 pF to 197.8 pF.

Question 11.
Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H. C = 80 μF, ft = 40 Q

(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across -the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating
Solution:
(a) Condition for resonance is when applied frequency matches with natural frequency.

(b) At resonance, impedance Z = R

(c) Potential drop across ‘L’

Question 12.
An LC circuit contains a 20 mH inductor and a 50 μF capacitor with initial charge of 10 mC. The resistance of the circuit in negligible. Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored

• completely electrical (i.e., stored in the capacitor)?
• completely magnetic (i.e., stored in the inductor)?

(d) At what times is the total energy shared equally between the inductor and capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Solution:

(a) Total energy is initially in the form of electric field within the plates of charged capacitor.

If we neglect the losses due to resistance of connecting wires, the total energy remain consumed during LC oscillations.
(b) Natural frequency of the circuit

(c) Instantaneous electrical energy

(d) timings for energy shared equally between inductor and capacitor.

(e) When a resistor is inserted in the circuit, eventually all the energy will be lost as heat across resistance. The oscillation will be damped.

Question 13.
A coil of inductance 0.50 H and resistance 100 Ω2 is connected to a 240 V. 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Solution:


(a) Virtual current in the coil

(b)

Question 14.
Obtain the answers (a) and (b) in Q. 13, if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Solution:
At very high frequency, X_L increases to infinitely large, hence circuit behaves as open circuit.

(a) Current in the coil, l_rms = 

This current is extremely small. Thus, at high frequencies, the inductive reactance of an inductor is so large that it behaves as an open circuit.
(b)

In dc circuit (after steady state), v = 0 and as such X_L = 0. In this case, the inductor behaves like a pure resistor as it has no inductive reactance.

Question 15.
A 100 μF capacitor in series with a 40 Q resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Solution:

(a) Virtual current in the coil

(b)

Question 16.
Obtain the answer to (a) and (b) in Q.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
Solution:
Given,

(a)

This value of current is same as that without capacitor in the circuit. So, at high frequency, a capacitor offer negligible resistance (0.1326 Ω in this case), it behave like a conductor.
(b)

In dc circuit, after steady state, v = 0 and accordingly, X_C = ∞, i.e., a capacitor amounts to an open circuit, i.e., it is a perfect insulator of current.

Question 17.
Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements of frequency. Source has emf 230 V and L = 5.0 H, C = 80 μF, ff = 40 Ω2.
Solution:

impedance of  R and X in parallel is given by

Thus, impedance Z = R and will be maximum. Hence, in parallel resonant circuit, current is minimum at resonant frequency. Current through circuit elements

Since, I_L and I_C are opposite in phase, so net current,

Question 18.
A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor.
(e) What is the total average power absorbed by the circuit? [‘Average’implies’averaged over one cycle’.
Solution:

(a) Inductive reactance , X_L = 2πƒL

(b) Potential drop across L, V_L = lυXL   X

(c) Average power transferred to inductor is zero, because of phase difference π/2

(d) Average power transferred to capacitor is also zero, because of phase difference π/2

(e) total power absorbed by the circuit

Question 19.
Suppose the circuit in Q. 18 has a resistance of 15 Ω2. Obtain the average power transferred to each element of the circuit and the total power absorbed.
Solution:
If the circuit has a resistance of 15 Ω2, now it is LCR series resonant circuit.

Question 20.
A series LCR circuit with L = 0.12 H, C = 480 μF, R = 23 Ω2 is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?
Solution:

(a) At resonant frequency, the current amplitude is maximum.

(b) Maximum power loss at resonant frequency, P = Eυlυ cos θ

(c) Let at an angular frequency, the source power is half the power at resonant frequency.

Question 21.
Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C= 27 μF, and R = 7.4 fl. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Solution:

we want to improve the quality factor to twice, without changing resonant frequency (without changing L and C).

Question 22.
Answer the following questions.
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across Cand the ac signal across L.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Solution:
(a) It is true that applied instantaneous voltage is equal to algebraic sum of instantaneous potential drop across each circuit element is series.

But the rms voltage applied is equal to vector sum of potential drop across each element, as voltage drops are in different phases.

(b) At the time of broken circuit of the induction coil, the induced high voltage charges the capacitor. This avoid sparking in the circuit.
(c) Inductive reactance, X_L = 2πƒL For a.c., X_c α ƒ

So, superimpose applied voltage will have all d.c. potential drop across Xc and will have most of a.c potential drop across XL.
(d) Inductor offer no hinderance to d.c. X_L = 0, so insertion of iron core does not effect the d.c. current or brightness of lamp connected. But it definitely effect a.c. current as insertion of iron core increases L = μm nl thus increases X_L (2πƒL). A.c. current in the E circuit reduces I_P =  and brightness of the bulb also reduces.
(e) A fluorescent tube is connected directly across a 220 V source, it would draw large current which may damage the filaments of the tube. So a choke coil which behaves as L-R circuit reduces the current to appropriate value, and that also with a lesser power loss.

An ordinary esistor used to control the current would have maximum power wastage as heat.

Question 23.
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Solution:

Question 24.
At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³s^-1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^-2).
Solution:
Work done by liquid pressure = pressure x volume shifted power of flowing water

Question 25.
A small town with a demand of 800 kW of 1 electric power at 220 V is situated 15 km away from an electric plant generating power at 440V. The resistance of the two wire line carrying power is 0.5 Q per km. The town gets 1 power from the line through a 4000-220 V step- down transformer at a sub station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterize the step up transformer at the plant.
Solution:

(a) Line power loss,

(b) Assuming no power loss due to leakage, total power need to be supply by the power plant

(c) Potential drop in the line,

Question 26.
Repeat the same exercise as in the previous question with the replacement of the earlier transformer by a 40,000-220 V step down transformer. (Neglect, as before, leakage losses through this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?
Solution:

(a) Line power loss,

(b) Power supplied by the plant

(c) Voltage drop in the line,

So, by supply of electricity at higher voltage, 40,000 V instead by 4000 V the power loss is reduced greatly that is why the electric power is always transmitted at very high voltage.

NCERT Exercises

Question 1.
Predict the direction of induced current in the situations described by the following Figures (a) to (f).



Solution:
Direction of induced current in all the situations shown above can be decided in the light of Lenz’s law.

Fig. (a) : South pole is moving closer, so the current is clockwise in the end of solenoid closest to magnet.
Fig. (b) : Following Lenz’s law, the current flow anticlockwise in the loop at the left and clockwise in the loop at the right.
Fig. (c) : Inner side of loop-1 become south pole whose strength increasing with increase in current. So the inner side of loop should also become south pole according to Lenz’s law.
Fig. (d) : Current is decreasing with increase in rheostat, so North pole is getting weaker, the current in inner part of loop-1 will flow clockwise.
Fig. (e) : Induced current in the right coil is from X to Y,
Fig. (f) : No induced current since magnetic lines of force are in the plane of the loop.

Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by figures.
(a) A wire of irregular shape turning into a circular shape:
(b) A circular loop being deformed into a narrow straight wire.

Solution:

(a) Due to change in shape, area increases and consequently magnetic flux linked with it also increases. Using Lenz’s law, an induced current is set up in the circular wire in the anticlockwise direction to produce opposing flux. So magnetic field due to it is directed upward.
(b) Due to deformation of circular loop into a straight wire, its area decreases and consequently magnetic flux linked with it decreases. So an induced current is set up in the -anticlockwise direction, hence magnetic field is upward.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Solution:
When the current changes through the solenoid, a change in magnetic field also take place within it. Initial magnetic field in solenoid,

Final magnetic field, B₂ = μ₀nI₂

Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the f e.m.f developed across the cut if velocity of loop is 1 cm s^-1 in a direction normal to the (a) longer side (b) shorter side of the loop? For how long does the induced voltage last in , each case?
Solution:
Here A = 2 × 2 = 16 cm² = 16 × 10 m², B = 0.3 T

Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Solution:
Constant and uniform magnetic field is parallel to axis of the wheel and thus normal to plane of the wheel.

Question 6.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^-1 in a uniform horizontal magnetic field of magnitude 3.0 × 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 ohm, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come,from?
Solution:
If the circular coil rotates in the magnetic field B at an angular velocity ot, then instantaneous induced emf can be calculated.


Source of the power is work done in rotating the coil.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^-4 Wb m^-2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Solution:
The direction of earth’s magnetic field is in the direction of geographical south to geographical north

Let us take a convenient way to represent all the directions.

(a) Instantaneous emf ε = Bυl

(b) Direction of emf. will be west to east.
(c) West end of the wire will be charged at higher potential.

Question 8.
Current in circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Solution:
Let ‘L’ is the coefficient of self inductance, the back emf

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Solution:
Let the current changes from 0 to 20 A in coil 1 and we are looking for change of flux linked with coil 2.

Question 10.
A jet plane is travelling towards west at a speed of 1800 km h^-1. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°.
Solution:
Earth magnetic field will have two components, B_H and B_v. It is vertical component which develop induced emf across the wing in N-S direction.

Question 11.
Suppose the loop shown in figure is stationary but the current feeding the electromagnet that produces the magnetic field is gradually ? reduced so that field decreases from its initial value of 0.3 T at the rate of 0.02 T s^-1. If the cut is joined and the loop has a resistance of 1.6Ω, how much power is dissipated by the loop as heat? What is the source of this power?

Solution:
Here area is constant but the magnetic field is reducing at a constant rate.

Source of the power is work done in changing magnetic field.

Question 12.
A square loop of side 12 cm with its sides ‘r parallel to X and Y axes is moved with a velocity of 8 cm s’1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^-3 T cm^-1 along the negative x:direction (that is it increases by 10“3 T cm-1 as one moves in the negative x-direction) and it is decreasing in time at the rate of 10^-3 T s^–1. r Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ
Solution:

Each side of square loop is 12 cm and magnetic field is decreasing along x direction.

Also the magnetic field is decreasing with time at constant rate

Induced emf and rate of change of magnetic flux due to only time variation

Induced emf and rate of change of magnetic flux due to change in position.

Both the induced emf have same sign and thus adds to provide net Induced emf in the loop

Question 13.
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of field region. Equivalently, one can give it quick 90° turn to bring its plane parallel to the field direction. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of coil and the galvanometer is 0.5 Ω. Estimate the field strength of magnet.
Solution:
Let the magnetic field between poles of loud speaker magnet is B.

Question 14.
Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mQ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s^-1 in the direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charges built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain
(d) What is the retarding force on the rod when If is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^-1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Solution:
Here rails, rod and magnetic field are in three mutually perpendicular directions.

(a) Switch K is open and rod moves with speed of 12 cm^-1 three mutually perpendicular directions. Induced emf/motional emf
ε = Bυl ε = 0.5 × 12 × 10^-2 × 15 × 10^-2 = 9 mV
(b) When the K is open, upper end of the rod become positively charge, and lower end become negatively charged.
When the K is closed the charge flows in closed circuit but the excess charge is maintained by the flow of charge in the moving rod under magnetic force.
(c) In the state when K is open very soon a stage is reached when force due to electric field which is due to potential difference induced balances the magnetic force on electrons. eE = Beυ  Motional emf V = Bυl.

(d) When the key is closed the current flows in a loop and the current carrying wire experience a retarding force in the magnetic field.

(e) To keep the rod moving in closed circuit at constant speed the force required is

when key K is open, no current flows and hence no retarding force, so no power is required to move at constant speed.
(f) Power lost in closed circuit due to flow of current

Power provided by external force to move the rod at constant speed is the source of this power lost.
(g) If B is parallel to rails, the induced/ motional emf will be zero.

Question 15.
An air cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^-3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Solution:
Magnetic field inside solenoid

Question 16.
(a) Obtain an expression for mutualin ductance , between a long straight wire and a square loop of side ‘a’ as shown in figure.

(b) Now assume that straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m s^-1. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1m and assume that loop has a large resistance.
Solution:
(a) As the magnetic field will be variable with distance from long straight wire, so the flux through square loop can be calculated by integration.

Let us assume a width ‘dr’ of the square loop at a distance ‘r’ from straight wire


(b) The square loop is moving right with a constant speed v, the instantaneous flux can be taken as

Question 17.
A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis. A uniform magnetic field extends over a circular region within the rim.  It is given by = o (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off?

Solution:
According to Faraday’s law of electromagnetic induction the induced emf is

Thus a relation between electric field and rate of change of flux can be established,

E exist along circumference of radius ‘a’ due to change in magnetic flux.

Linear charge density on rim is A. So, total charge on rim Q = λ2πa …(ii) Electric Force on the charge

NCERT Exercises

Question 1.
Answer the following questions regarding earth’s magnetism:
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 10²² J T^-1 located at its centre. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
Solution:
(a) Angle of declination (magnetic declamatory) θ, angle of dip δ, horizontal component of earth’s magnetic field B_H are the quantities which are considered as elements of earth’s magnetic field.
(b) Britain is closer to magnetic north pole hence angle of dip is much larger, nearly 70° in Britain.
(c) Melbourne is closer to south pole, so north of the assumed magnet buried within earth lies inside, hence the field lines would seem to be coming out of the ground.
(d) At geomagnetic north or south pole, angle of dip is 90°, where horizontal component of earth’s magnetic field B_H is zero. A compass needle can only turn in horizontal plane, so it can point in any direction as B_H = 0, which governs its direction.
(e) Let us consider the magnetic field on surface of earth due to assumed bar magnet of dipole moment 8 × 10²² J T^-1 located at centre of earth. The magnetic field at point P equatorial position on earth can be calculated as

(f) Localised magnetic dipoles can develop due to magnetised mineral deposits or movement of charged ions in atmosphere.

Question 2.
Answer the following questions:
(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the battery (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30000 km). What agencies may be responsible for this distortion?
(f) Interstellar space has an extremely weak magnetic field of the order of 10^-12 T. Can such a weak field be of any significant consequence? Explain.
Solution:
(a) Yes, earth’s field undergoes a change with time. For e×ample, daily changes, annual changes, secular changes with period of the order of 960 years and irregular changes like magnetic storms. Time .scale for appreciable change is roughly a few hundred years.
(b) The earth’s core does contain iron but in the molten form only. This is not ferromagnetic and hence it cannot be treated as a source of earth’s magnetism.
(c) One of the possibilities is the radioactivity in the interior of the earth. But it is not certain.
(d) Earth’s magnetic field gets recorded weakly in certain rocks during their solidification. An analysis of these rocks may reveal the history of earth’s magnetism. The earth’s magnetic field gets modified by the field produced by motion of ions in earth’s ionosphere.
(f) When a charged particle moves in a magnetic field, it is deflected along a circular path such that

when B is low, R is high i.e. radius of curvature of path is very large. Therefore, over the gigantic interstellar distance, the deflection of charged particles becomes less noticeable.

Question 3.
A short bar magnet placed with its a×is at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^-2 J. What is the magnitude of magnetic moment of the magnet?
Solution:
Torque x = MB sinθ 4.5 × 10^-2 = M (0.25 sin 30°) Magnetic dipole moment, M = 0.36 J T^-1.

Question 4.
A short bar magnet of magnetic moment M = 0.32 J T^-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Solution:
(a) An equilibrium is stable, if on disturbing the magnet, it comes back to same initial state. Bar magnet is in stable equilibrium at 0 = 0° Potential energy.


(b) A bar magnet is in unstable equilibrium, if on disturbing from its position, it further gets disturb and do not come back to previous position of equilibrium. At 0 = 180°, the equilibrium is unstable. Potential energy

Question 5.
A closely wound solenoid of 800 turns and area of cross section 2.5 × 10^-4 m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Solution:

A current carrying closely wound solenoid acts like bar magnet. Each of the turn provide a dip’ole moment and all turns together provides the dipole moment of the magnet. Total magnetic moment, M = NIA = 800 × 3 × 2.5 × 10^-4 = 0.6 A m²

Question 6.
If the solenoid in the previous question is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its a×is makes an angle of 30° with the direction of applied field?
Solution:
The solenoid behaves as a bar magnet

Question 7.
A bar magnet of magnetic moment 1.5 J T^-1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:

• normal to the field direction,
• opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?
Solution:
(a) Work required to turn the dipole

(b) Torque when 0 = 90°

Question 8.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 T is set up at an angle of 30° with the axis of the solenoid?
Solution:
(a) Magnetic moment associated with solenoid
M = NIA = 2000 × 4 × 1.6 × 10^-4 = 1.28 A m²
(b) Force on the solenoid will be zero in uniform magnetic field.
Torque x = MB sinθ = 1.28 × 7.5 × 10^-2 × sin 30° or × = 4.8 × 10^-2 N m The torque tends to align the solenoid in the direction of magnetic field.

Question 9.
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10^-2 T.The coil is free to turn about an a×is in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^-1. What is the moment of inertia of the coil about its axis of rotation?
Solution:
The circular coil carries a dipole moment

Question 10.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Solution:
North tip is pointing down at 22° with horizontal, hence the location is in northern hemisphere.

Question 11.
At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Solution:
Compass needle points 12° west of geographical north, hence angle of declination 0 is 12° west. North tip of magnetic needle is 60° above horizontal, hence the location is in southern hemisphere and angle of dip is 60°. Magnitude of net magnetic field can be calculated as

Question 12.
A short bar magnet has a magnetic moment of 0.48 J T^-1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Solution:

Question 13.
A short bar magnet placed in a horizontal plane has its a×is aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Solution:
Angle of dip is zero at the place, so earth’s magnetic field is uniform with magnitude 0.36 G in the direction geographi¬cal south to geographical north.

Question 14.
If the bar magnet in the previous problem is turned around by 180°, where will the new null points be located? wairii When the bar magnet is turned through 180°, neutral points would lie on equatorial line, so that
Solution:
when the bar magnet is tuned through 180, neutral points would lie on equatorial line, so that

Question 15.
A short bar magnet of magnetic moment 5.25 × 10^-2 J T^-1 is placed with its a×is perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on
(a) its normal bisector and
(b) its a×is. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Solution:

Question 16.
Answer the following questions:

1. Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
2. Why is diamagnetism, in contrast, almost independent of temperature?
3. If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
4. Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
5. Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?
6. Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

Solution:

1. In paramagnetics, the tendency to disrupt the alignment of molecular dipoles with the external magnetising field arising from random thermal motion is reduced at lower temperatures.
2. In diamagnetics, the molecular dipole moments always align in direction opposite to that of external magnetising field, inspite of the internal motion of atoms.
3. As bismuth is diamagnetic, so the field in the toroid with bismuth core will be slightly less than when the core is empty.
4. No, the permeability of a ferromagnetic material is not independent of the magnetic field. It is more at higher fields.
5. As the magnetic permeability p of a ferromagnet is much larger than unity i.e., p » 1, so magnet field lines are always nearly normal to the surface of a ferromagnet at every point.
6. Yes, but for the maximum possible magnetisation of paramagnetic sample, impractically very high magnetising fields are required.

Question 17.
Answer the following questions:

1. Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
2. The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through > repeated cycles of magnetisation, which piece will dissipate greater heat energy?
3. A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory? E×plain the meaning of this statement.
4. What kind of ferromagnetic material is used for coating magnetic tapes in a ‘cassette player, or for building ‘memory stores’in modern computer?
5. A certain region of space is to be shielded from magnetic fields. Suggest a method.

Solution:

1. In a specimen of a ferromagnets, the atomic dipoles are grouped together in domains. All the dipoles of a domain are aligned in the same direction and have net magnetic moment. In an unmagnetised substance these domains are randomly distributed so that the resultant magnetisation is zero. When the substance is placed in an external magnetic field, these domains align themselves in the direction of the field. Some energy is spent in the process of alignment when the external field is removed, these domains ‘ do not come back into their random positions completely. The substance retains some magnetisation. The energy spent in the process of magnetisation is f’ not fully recovered. The balance of energy is lost as’ heat. This is the basic cause for irreversibility of the magnetisation curve of a ferromagnetic substance.
2. Carbon steel piece, because the heat produced in complete cycle of magnetisation is directly proportional to he area under the hysteresis loop.
3. Magnetisation of a ferromagnet is not a single valued function of the magnetising ‘ field. Its value for a particular field depends both on the magnetising field and on the history of its magnetisation i.e. how many cycles of magnetisation it has gone through etc. So, the value of magnetisation is a record or memory of its cycles of magnetisation. If information bits can be made to correspond to these cycles, the system displaying such a hysteresis loop can act as a device for storing information.
4. Ferrites or ceramics which is specially treated barium iron oxides.
5. By surrounding the region with soft iron rings, as magnetic field lines will be drawn into the rings and the enclosed space becomes free of magnetic field.

Question 18.
A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north df east. The magnetic meridian of the place happens to be 10° west of ” the geographic meridian.The earth’s magnetic dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Solution:

The neutral point can be achieved at a location above cable, where magnetic field of cable is balanced by earth’s magnetic field B_H.

Question 19.
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of- dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at point 4.0 cm below the cable?
Solution:
Let us first decide the directions which can best represent the situation.

Telephone cable carry a total current of 4.0 A in direction east to west. We want resultant magnetic field 4.0 cm below.

Question 20.
A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian when the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of earth’s magnetic field at the location.
(b) The current in the coil is reversed and the coil is rotated about its vertical a×is by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Solution:
Here, n = 30, r = 12 cm = 12 × 10^-2 m. i = 0.35 A ,H=? As is clear from figure shown the needle can point west to east only when H = B sin 45°


(b) When current in coil is reversed and coil is turned through 90° anticlockwise, the direction of needle will reverse (i.e. it will point from east to west). This follows from figure shown.

Question 21.
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10^-2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?
Solution:
The magnetic dipole experiences torque due to both the fields and is in equilibrium.

Question 22.
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.4 G normal to the initial direction. Estimate the up or dowri deflection of the beam over a distance of 30 cm (m_e = 9.11 × 10^-31 kg). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Solution:
Kinetic energy of electron

Velocity in x direction remains constant, hence time to cross 30 cm

A magnetic force F = euB is acting in vertical direction, which provides acceleration in vertical direction

Question 23.
A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10^-23 J T^-1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)
Solution:
Dipole moment at complete saturation

Question 24.
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Solution:
Magnetic field B in the core

Question 25.
The magnetic moment vectors  and  associated with the intrinsic spin angular momentum   and orbital angular momentum   respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by  =   and  =  Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Solution:
Out of the two relations given, only one is in accordance with classical physics. This is

where r is the radius of the circular orbit, which the electron of mass m and charge (- e) completes in time T. Divide (i) by (ii),

of quantum mechanics.