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NCERT Class 12 Physics Textbook Chapter 6 With Answer PDF Free Download
Chapter 6: Electromagnetic Induction
6.1 Introduction
Electricity and magnetism were considered separate and unrelated phenomena for a long time.
In the early decades of the nineteenth century, experiments on electric current by Oersted, Ampere, and a few others established the fact that electricity and magnetism are inter-related.
They found that moving electric charges produce magnetic fields. For example, an electric current deflects a magnetic compass needle placed in its vicinity.
This naturally raises the questions like: Is the converse effect possible? Can moving magnets produce electric currents? Does nature permit such a relation between electricity and magnetism?
The answer is a resounding yes! The experiments of Michael Faraday in England and Joseph Henry in the USA, conducted around 1830, demonstrated conclusively that electric currents were induced in closed coils when subjected to changing magnetic fields.
In this chapter, we will study the phenomena associated with changing magnetic fields and understand the underlying principles. The phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction.
Magnetic Flux
Faraday’s great insight lay in discovering a simple mathematical relation to explaining the series of experiments he carried out on electromagnetic induction.
However, before we state and appreciate his laws, we must get familiar with the notion of magnetic flux, Φ B.
Magnetic flux is defined in the same way as electric flux is defined in Chapter 1. Magnetic flux through a plane of area A placed in a uniform magnetic field B (Fig. 6.4) can be written as
Φ B = B . A = BA cos θ (6.1)
where θ is angle between B and A. The notion of the area as a vector has been discussed earlier in Chapter 1. Equation (6.1) can be extended to curved surfaces and non-uniform fields.
If the magnetic field has different magnitudes and directions at various parts of a surface as shown in Fig. 6.5, then the magnetic flux through the surface is given by
where ‘all’ stands for summation over all the area elements dAi comprising the surface and Bi
is the magnetic field at the area element dAi.
The SI unit of magnetic flux is weber (Wb) or tesla meter squared (T m2). Magnetic flux is a scalar quantity.
6.4 Faraday’s Law Of Induction
From the experimental observations, Faraday arrived at a conclusion that an emf is induced in a coil when magnetic flux through the coil changes with time.
Experimental observations discussed in Section 6.2 can be explained using this concept. The motion of a magnet towards or away from coil C1 in Experiment 6.1 and moving a current-carrying coil C2 towards or away from coil C1 in Experiment 6.2, change the magnetic flux associated with coil C1. The change in magnetic flux induces emf in coil C1.
It was this induced emf that caused an electric current to flow in coil C1 and through the galvanometer.
A plausible explanation for the observations of Experiment 6.3 is as follows: When the tapping key K is pressed, the current in coil C2 (and the resulting magnetic field) rises from zero to a maximum value in a short time.
Consequently, the magnetic flux through the neighboring coil C1 also increases. It is the change in
magnetic flux through coil C1 that produces an induced emf in coil C1.
When the key is held pressed, the current in coil C2 is constant. Therefore, there is no change in the magnetic flux through coil C1, and the current in coil C1 drops to zero.
When the key is released, the current in C2 and the resulting magnetic field decrease from the maximum value to zero in a short time.
This results in a decrease in magnetic flux through coil C1 and hence again induces an electric current in coil C1.
The common point in all these observations is that the time rate of change of magnetic flux through a circuit induces emf in it. Faraday stated experimental observations in the form of a law called Faraday’s law of electromagnetic induction.
6.5 LENZ’S LAW AND CONSERVATION OF ENERGY
In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced a rule, known as Lenz’s law which gives the polarity of the induced emf in a clear and concise fashion.
The statement of the law is: The polarity of induced emf is such that it tends to produce a current
which opposes the change in magnetic flux that produced it.
The negative sign shown in Eq. (6.3) represents this effect. We can understand Lenz’s law by examining Experiment 6.1 in Section 6.2.1.
In Fig. 6.1, we see that the North-pole of a bar magnet is being pushed towards the closed coil. As the North-pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases.
Hence current is induced in the coil in such a direction that it opposes the increase in flux.
This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet.
Note that the magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet.
Similarly, if the Northpole of the magnet is being withdrawn from the coil, the magnetic flux
through the coil will decrease.
To counter this decrease in magnetic flux, the induced current in the coil flows in a clockwise direction and its Southpole faces the receding North-pole of the bar magnet.
This would result in an attractive force that opposes the motion of the magnet and the corresponding decrease in flux.
| Author | NCERT |
| Language | English |
| No. of Pages | 29 |
| PDF Size | 934 KB |
| Category | Physics |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction
Q 2. We are rotating a 1 m long metallic rod with an angular frequency of 400 red s^{-1}s−1 with an axis normal to the rod passing through its one end. And on to the other end of the rod, it is connected with a circular metallic ring. There exist a uniform magnetic field of 0.5 T which is parallel to the axis everywhere. Find out the emf induced between the center and the ring.
Ans:
Length of the rod = 1m
Angular frequency,\omega = 400 \; rad/sω=400rad/s
Magnetic field strength, B = 0.5 T
At one of the ends of the rod, it has zero liner velocity, while on its other end it has a linear velocity of I \omegaIω
Average linear velocity of the rod, v = \frac{I \omega + 0}{ 2 } = \frac{I \omega}{2}v=2Iω+0=2Iω
Emf developed between the centre and ring.\\e = Blv = Bl\left ( \frac{I \omega }{2} \right ) = \frac{B l^{2 } \omega}{2} \\ \\ = \frac{0.5 \times \left ( 1 \right )^{2} \times 400}{2} = 100 Ve=Blv=Bl(2Iω)=2Bl2ω=20.5×(1)2×400=100V
Hence, the emf developed between the center and the ring is 100 V.
Q 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?
Ans:
Number of turns on the solenoid – 15 turns / cm = 1500 turns / m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2
The current carried by the solenoid changes from 2 A to 4 A.
Therefore, Change in current in the solenoid, di = 4 – 2 = 2 A
Change in time, dt = 0.1 s
According to Faraday’s law, induced emf in the solenoid is given by:e = \frac{d\phi }{dt} \;\;\;\;\;\;\;\;\; . . . (1)e=dtdϕ…(1)
Where,\phiϕ = Induced flux through the small loop
= BA . . . (2)
B = Magnetic field
=\mu _{0} niμ0ni\mu _{0}μ0 = Permeability of free space
= 4\pi \times 10 ^{-7} \; H/m4π×10−7H/m
Hence, equation (1) can be reduced to:e = \frac{d}{dt}\left ( BA \right )e=dtd(BA)e = A\, \mu _{0}\, n \times \left (\frac{di}{dt} \right ) \\ \\ = 2 \times 10^{-4} \times 4 \pi \times 10 ^{-7} \times 1500 \times \frac{2}{0.1}\\ \\ = 7.54 \times 10^{-6}\; Ve=Aμ0n×(dtdi)=2×10−4×4π×10−7×1500×0.12=7.54×10−6V
Hence, the induced voltage in the loop is 7.54 \times 10^{-6}\; V7.54×10−6V
Q 4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the 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?
Ans:
Length of the wired loop, l = 8 cm = 0.08 m
Width of the wired loop, b = 2 cm = 0.02 m
Since, the loop is a rectangle, area of the wired loop,
A = lb
= 0.08 × 0.02
= 16 \times 10 ^{-4} \, m^{2}16×10−4m2
Strength of magnetic field, B = 0.3 T
Velocity of the loop, v = 1 cm / s = 0.01 m / s
(i) Emf developed in the loop is given as:
e = Blv
= 0.3 × 0.08 × 0.01 =2.4\times 10^{-4} \; V2.4×10−4V
Time taken to travel along the width ,t = \frac{Distance \; travelled}{Velocity} = \frac{b}{v} \\ \\ = \frac{0.02}{0.01} = 2st=VelocityDistancetravelled=vb=0.010.02=2s
Hence, the induced voltage is 2.4\times 10^{-4} \; V2.4×10−4V which lasts for 2 s.
(ii) Emf developed, e = Bbv
= 0.3 × 0.02 × 0.01 =0.6 \times 10^{-4} \, V0.6×10−4V
Time taken to travel along the length, t = \frac{Distance \; travelled}{Velocity} = \frac{l}{v} \\ \\ = \frac{0.08}{0.01} = 8st=VelocityDistancetravelled=vl=0.010.08=8s
Hence, the induced voltage is 0.6 \times 10^{-4} \, V0.6×10−4V which lasts for 8 s.
Q 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 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Ans:
Maximum emf induced = 0.603 V
Average emf induced = 0 V
Maximum current in the coil = 0.0603 A
Power loss (average) = 0.018 W
(Power which is coming from external rotor)
Circular coil radius, r = 8 cm = 0.08 m
Area of the coil,A = \pi r^{2} = \pi \times \left ( 0.08 \right )^{2}m^{2}A=πr2=π×(0.08)2m2
Number of turns on the coil, N = 20
Angular speed,\omega = 50 \; rad/sω=50rad/s
Strength of magnetic,B = 3 \times 10 ^{-2} \; TB=3×10−2T
Total resistance produced by the loop,R = 10 \OmegaR=10Ω
Maximum emf induced is given as:e = N\omega \; AB\\ = 20 \times 50 \times \pi \times \left ( 0.08 \right )^{2} \times 3 \times 10^{-2} \\ = 0.603 \; Ve=NωAB=20×50×π×(0.08)2×3×10−2=0.603V
The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero.
Maximum current is given as:I = \frac{e}{R} \\ = \frac{0.603}{10} = 0.0603\; AI=Re=100.603=0.0603A
Average power because of the Joule heating:P = \frac{el}{2} \\ = \frac{0.603 \times 0.0603}{2} = 0.018\; WP=2el=20.603×0.0603=0.018W
The torque produced by the current induced in the coil is opposing the normal rotation of the coil. To keep the rotation of the coil continuously, we must find a source of torque that opposes the torque by the emf, so here the rotor works as an external agent. Hence, dissipated power comes from the external rotor.
NCERT Class 12 Physics Textbook Chapter 6 With Answer PDF Free Download