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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.

NCERT Exercises

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.4 A. What is the magnitude of the magnetic field 6 at the centre of the coil?
Solution:
The magnetic field at the centre of a circular coil having 100 turns.

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the field 6 at a point 20 cm from the wire?
Solution:
Magnetic field due to a long straight wire

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Solution:
Let us first decide the standard directions on the plane of paper.

Magnitude of magnetic field

Question 4.
A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Solution:
The standard directions on the plane of paper can be different according to requirements.

Direction of magnetic field can be observed by right hand palm rule and it is southward.

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Solution:

Let l be length of wire, carrying a current of 8 A at an angle 30° with the magnetic field. Force on the wire, F = |B| sinθ Force per unit length F/l = |B| sinθ F/l = 8 × 0.15 × sin 30° = 0.6 N m^-1.

Question 6.
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Solution:
The magnetic field inside the solenoid is along its axis. the current in the wire flows perpendicular to the axis. F = |B| sin 90° or F = 10 × 0.27 × 3 × 10^-2 x 1 = 0.081 N

Question 7.
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Solution:
Force of attraction per unit length on two parallel wires carrying current in same direction.

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current the carried is 8.0 A, estimate the magnitude of 8 inside the solenoid near its centre.
Solution:
Total number of turns in 80 cm length of solenoid can be calculated

Question 9.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Solution:

Torque experienced by the coil carrying current in the given magnetic field.

Question 10.
Two moving coil meters, M₁ and M₂ have following particulars: R₁ = 10Ω, N₁ = 30, A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10^-3 m², B₂ = 0.50 T. (The spring constants are identical for two meters). Determine ratio of (i) current sensitivity (ii) voltage sensitivity of M2 and A4,.
Solution:
Current sensitivity of a moving coil galvanometer is defined as

(ii) Ratio of voltage sensitivity

Question 11.
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^-4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e= 1.6 × 10^-19 C,m_e = 9.1 × 10^-31 kg)
Solution:
The magnetic force ƒ = quB act normal to the direction of motion, thus provide the necessary centripetal force to follow the circular path.

Question 12.
In question 11, obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Solution:
Time period can be calculated.

The frequency of revolution of electron is independent of speed of electron.

Question 13.
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? [All other particulars are also unaltered].
Solution:
(a) The given coil is circular and is suspended such that field lines makes angle 60° with normal of the coil.A similar torque is required to prevent the coil from turning.

A similar torque is required to prevent the coil from turning.
(b) As long as the area of the planar coil remains same, the torque on the coil is also same, irrespective of the shape.

Question 14.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Solution:
The concentric coils are in the plane north to south. Let us decide the directions with conveniences in the plane of paper.

Magnetic field at centre due to coil X

So, a magnetic field nearly 1.6 × 10^-3 T will appear at centre of the circular coil directed towards west.

Question 15.
A magnetic field of 100 G (1G = 10^-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^3-3 m². The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m”1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Solution:
The magnetic field required is 100 G in a length of 10 cm. So, the solenoid should have a length larger than 10 cm. The maximum current capacity of wire is 15 A, so less than 15 A should flow in wire. Now using

So, one combination for the desired 100 G magnetic field in a length 10 cm can be a total length of solenoid of 50 cm and current in the solenoid of 10 A and total turns nearly 400.

Question 16.
Fora circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its

(a) Show that this reduces to the familiar result for field at the centre of the coil
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Solution:
On the axis of the circular coil of radius R and N turns carrying a current I at a distance x from centre the magnetic field is

Which is the same result as is obtained by integration at centre of the coil.
(b) The side of the loop where current flows clockwise becomes south pole and where the current appear anticlockwise becomes north pole. Direction of magnetic field is from south pole to north pole.

Now, two parallel coils each of radius R with N turns. The magnetic field of both the coils add.

Question 17.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field
(a) outside the toroid,
(b) inside the core of the toroid, and
(c) in the empty space surrounded by the toroid.
Solution:

Total turns in the toroid are given 3500.
(a) outside the toroid the magnetic field is zero.
(b) Inside the core of toroid, the magnetic field will be

(c) Magnetic field in the empty space surrounded by the toroid is also zero.

Question 18.
Answer the following questions:
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Solution:
(a) If a charged particle move parallel or anti parallel to the magnetic field, no magnetic force will act on it and it move undeflected. So, in the given condition either the charged particle enters east to west or west to east.
(b) A magnetic force can only change the direction of charged particle but never changes magnitude of speed as force act normal to direction of speed. So, charged particle may follow a complicated trajectory, but its speed remains the same.
(c)

If,we want the electron to move undeflected in the presence of electric and magnetic fields, then the electric force should be balanced by magnetic force.

So, the magnetic field should act in downward direction.

Question 19.
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Solution:
A electron which is accelerated by a potential difference of 2.0 kV will have a kinetic energy gained 2000 eV.

(a) When the electron enters in the uniform r magnetic field which is normal to the velocity of electron the electron follows a circular path r of radius.

(b) When the magnetic field makes an angle 30° with the initial velocity, the trajectory of the electron becomes helical. radius of the helical path is

Question 20.
A magnetic field set up using Helmholtz coils (described in question 16 above) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10^-5 V m^-1, make a simple guess as to what the beam contains. Why is the answer not unique?
Solution:
Narrow beam of charged particles remains undeflected and is perpendicular to both electric field and magnetic fields which are mutually perpendicular. So, the electric force is balanced by magnetic force. qE = quB Speed of charged particles

Because the beam is accelerated through 15 kV, if charge is q, then kinetic energy gained by charged particles

Here, we can only obtain charge to mass ratio and same ratio can be in Deuterium ions, He⁺⁺, Li⁺⁺⁺, so the beam can contain any of these charged particles.

Question 21.
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before?
[Ignore the mass of the wires.] g = 9.8 m s^-2
Solution:

Tension in the strings and magnetic force |B| balance the weight of wire. 2 T + |B| = mg

Question 22.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Solution:
Two wires connecting battery of an automobile to the starting motor carry 300 A current in opposite direction. So, the force is repulsive between them. Force per unit length

Question 23.
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast- northwest direction.
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Solution:

The magnetic field is in the direction east to west and in the cylindrical region of radius 10 cm.
(a) A current carrying wire, intersects the axis. Force on wire, F = |B| = IB (2r) = 7 × 1.5 × 2 × 1 × 10^-2 F = 2.1 N Downwards
(b) By turning the wire by an angle 45° in NE and NW direction the force remain the same,/= 2.1 N Downwards
(c) Now the wire is lowered from the axis by a distance of 6.0 cm.

Length of wire inside cylindrical region is 16 cm now. So, the force is F = |B| = 7 × 1.5 × 16 × 10^-2 = 1.68 N, vertically downwards

Question 24.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in figure? What is the force on each case? Which case corresponds to stable equilibrium?

Solution:
(a) Let us detail each case separately,


So, torque is 1.8 × 10^-2 N-m along -y direction net force on the coil is zero coil not in equilibrium.
(b) Dipole moment is along +x direction

So, torque is 1.8 x 10-2 N m along -y direction. Net force on the coil is zero, coil is not in equilibrium
(c) Dipole moment is along -y direction

Net force on the coil is zero, coil is not in equilibrium.
(d)

Dipole moment is at an angle 150° with the +x direction.

At an angle 240° with the +x direction net force on the coil is zero, coil is not in equilibrium.

Negative energy shows equilibrium is stable,
(f) Dipole moment is along – z direction

Positive energy shows equilibrium is unstable.

Question 25.
A circular coil of 20 turns and radius 10 cm is place in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^-5 m², the free electron density in copper is given to be about 10²⁹   m-³.)
Solution:
The magnetic field is normal to the plane of the coil, so condition of minimum torque.

(a) Torque on the coil τ = NIB A sinθ here θ = 0° τ = 0
(b) Force on every element of the coil is cancelled by force on corresponding element. Net force on the unit is zero.

(c) To calculate force on each electron, let us find drift velocity.

Question 26.
A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside solenoid (near its centre) normal to its axis; both the wire and axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current [with appropriate sense of circulation] in the windings of the solenoid can support the weight of the wire? g = 9.8 m s^-2.
Solution:
Magnetic in the solenoid B = μ_0nI

A current carrying wire is suspended inside solenoid and a current 7 = 6.0 A is flowing in it. To balance the weight of the wire by the magnetic force

Question 27.
A galvanometer coil has a resistance of 12 Q and the metre shows full scale deflection for a current of 3 mA. How will you convert metre into a voltmeter of range 0 to 18 V?
Solution:
By using the formula

We can calculate required resistance to be connected in series.

Question 28.
A galvanometer coil has a resistance of 15 Q and the meter shows full scale deflection for a current of 4 mA. How will you convert metre into an ammeter of range 0 to 6 A?
Solution:
For converting galvanometer into ammeter of required range, required shunt can be calculated

NCERT Exercises

Question 1.
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
Solution:
The maximum current will be obtained when no external resistance is offered by wire joining the two terminals.

Question 2.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Solution:

Question 3.
(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Solution:
(a) Total resistance of the combination in series

Question 4.
(a) Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance. Determine the current through each resistor, and the total current drawn from the battery.
Solution:
(a) Total resistance of the combination in parallel


(b) Potential of 20 V will be same across each resistor, so current

Question 5.
At room temperature (27.0°C) the resistance of a heating element is 100 O. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^{3 0}C^-1.
Solution:

Question 6.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^-7 m₂, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Solution:

Question 7.
A silver wire has a resistance of 2.1 Ω, at 27.5°C and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.
Solution:

Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C ?Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^-4 °C^-1.
Solution:
At room temperature 27 °C, the resistance of the heating element.

Question 9.
Determine the current in each branch of the network shown in figure.

Solution:
Marti Let us first distribute the current in different branches. Now, eΩ actions for different loops using Kirchhoff’s II^nd law,



Question 10.
(a) In a meter bridge as shown in figure, the balance point is found to be at 39.5 cm from the end A, when the resistor is of 12.5 Ω. Determine the resistance of x. Why are the connection between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if x and are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Solution:
(a)

The connections are made of thick copper strips so as to provide negligible resistance by connecting wires.
(b) If x and Y are interchanged the balance point will be at 60.5 cm from A.
(c) In balanced condition of the bridge, the cell and the galvanometer can be exchanged, the galvanometer will still show zero deflection.

Question 11.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Solution:

Now, terminal voltage of the battery during charging
V = E + lr = 8 + 7(0.5) = 11.5 V A series resistance is joined in the charging circuit to limit the excessive current so that charging is slow and permanent.

Question 12.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Solution:
The potential gradient remains the same, as there is no change in the setting of standard circuit.

Question 13.
The number density of free electrons in a copper conductor estimated is ,8.5 × 10²⁸ m^-3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^-6 m² and it is carrying a current of 3.0 A.
Solution:
We can first calculate drift velocity of the electrons from the given data I = Aneu_d

Question 14.
Theearth’s surface has a negative surface charge density of 10^-9  cm^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10⁶ m).
Solution:
Due to negative charge on earth, an electric field is into the earth surface due to which the positive ions of atmosphere are constantly pumped in and an equivalent current of 1800 A is established across the globe. Let us first calculate total negative charge on earth

Question 15.
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance of 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?
Solution:
(a) Six cells are joined in series.

Equivalent emf is 2 × 6 = 12 V
Equivalent internal resistance is 0.015 × 6 = 0.09 Ω
Current drawn from supply

(b) Maximum current is drawn from a battery when external resistance is tested to be zero

To start a car, a current of the order of 100 A is needed, so the battery mentioned above can not drive the starting motor.

Question 16.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables.

Solution:
Two wires have same length, and resistance. As the specific resistances are unequal, the areas are different. For copper wire, R_cu = p_cu 

Thus, the aluminium wire for the same resistance is very light than copper and that is why aluminium wires are preferred for overhead power cables.

Question 17.
What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Solution:
We can find resistance of the alloy manganin for all the readings as follows :

Here we can conclude that Ohm’s law is valid to a high accuracy and resistance of the alloy is nearly constant at all currents.

Question 18.
Answer the following questions:

1. A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
2. Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
3. A low voltage supply from which one needs high currents must have very low internal resistance. Why?
4. A high tension (HT) supply of, say, 6 kV must have a large internal resistance. Why?

Solution:

1. The current will be constant because it is given to be steady.
   
2. Ohm’s law is not a fundamental law in nature. It is not universally followed. Semiconductor diodes, transistors, their mistors, vacuum tubes etc. do not follow Ohm’s law.
3. If emf of supply battery is E and internal resistance r, then current through an external resistance R is given by  so, internal resistance r should be least to supply high currents.
4. In high tension supply, the internal resistance is made large. Because, if accidently the short circuiting take place, the excessive current produced should not cross the safety limits.

Question 19.
Choose the correct alternative:

1. Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
2. Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
3. The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
4. The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022 or 103).

Solution:

1. Alloys of metals usually have greater resistivity than that of their constituent metals.
2. Alloys usually have much lower temperature coefficients of resistance than pure metals.
3. The resistivity of the alloy manganin is nearly independent of increasing temperature.
4. The resistivity of a typical insulator (e.g. amber) is greater than that of a metal by factor of the order of 1022.

Question 20.
(a) Given n resistors each of resistance R, how will you combine them to get the

• maximum
• minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given the resistances of 1 Ω, 2Ω, 3Ω, how will be combine them to get an equivalent resistance of

• (11/3) Ω
• (11/5) Ω,
• 6 Ω,
• (6/11) Ω?

(c) Determine the equivalent resistance of network shown in figure.

Solution:
(a)

• For maximum effective resistance, all the resistors should be joined in series. Rm_max = R + R + R+……..n or R_max = nR
• For minimum effective resistance, all the resistors should be joined in parallel.

(b) All possible combinations with resistances 1 Ω, 2 Ω and 3 Ω are


So, the combinations for the desired results can be selected.

Question 21.
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in figure. Each resistor has 1 Ω resistance.

As shown in diagram the loop of resistance shown by arrow is repeated at times, let us assume the equivalent resistance is x, so by adding one more loop the resistance will remain as x.

Question 22.
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf E and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

1. What is the value of E?
2. What purpose does the high resistance of 600 kΩ have?
3. Is the balance point affected by this high resistance?
4. Is the balance point affected by the internal resistance of the driver cell?
5. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
6. Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Solution:

1. For emf of 1.02 V, the balance point is 67.3 cm of wire. For unknown emf E, the balance point is 82.3 cm of wire.
   
2. A high resistance of 600 kΩ is joined in series with cell so as to prevent the galvanometer from excessive current, when the jockey is touched far from null point. Once the null point is closely known the 600 kΩ is short circuited so that exact position of null point can be achieved.
3. In presence of high resistance we will obtain null point for a longer length then a sharp position, it is because of very low current near null point.
4. Internal resistance of the cell do not affect the balance point, as no current is drawn at null point.
5. If the driver cell of standard circuit has an emf smaller than emf of the cell to be balanced then null point will not be achieved in the length of potentiometer wire and galvanometer will provide only one side deflection always.
6. In the given state, the circuit will be unsuitable, as for E of the order of mV the null point will be very close to end A and percentage error in the measurement of emf will be larger.
   To make suitable arrangement the potential drop in potentiometer wire is brought down to few mV, so that balance point is obtained at suitable length and this is done by introducing a high resistance in series in the standard calibration circuit.
   

Question 23.
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance x is 68.5 cm. Determine the value of x. What might you do if you failed to find a balance point with the given cell of emfε?

Solution:
For comparison of resistances, R and x should be in series and same current should flow through them.

If we fail to find balance point with cell of e.m.f. e, in that case we should reduce the current I and for that e.m.f. E of cell responsible for current I should reduce, or a resistor can also be attached in series with R and x to reduce current.

Question 24.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Solution:
In the open circuit, the balance point is obtained for the emf of 1.5 V.

When the external circuit is connected, a current is drawn from the cell of 1.5 V in external resistance of 9.5 E1. Now the balance point is obtained for terminal potential