NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction
Q. Differentiate between self-induction and mutual induction.
Self Induction | Mutual Induction | |
(i) | When the current flowing through a coil changes then induced emf is produced in the coil itself. This phenomenon is called selfinduction. |
When current in one of the two closely lying coils is changed, the emf is induced in the second coil. This phenomenon is called mutual induction. |
(ii) | Phenomenon of self-induction is characteristic of single coil. | It is a characteristic of two coil. |
(iii) | The induced current is produced in the same coil in which main current is flowing. Hence, the induced current affects the main current. |
Main current flows in one coil and induced current is produced in the second coil. Hence, main current is not directly affected. |
- (i) What would you do to obtain a large deflection of the galvanometer ?
- (ii) How would you demonstrate the presence of an induced current in the absence of a galvanometer ?
- Ans. (i) To obtain a large deflection, one or more of the following steps can be taken :
- (a) Use a rod made of soft iron inside the coil C_{2}.
- (b) Connect the coil to a powerful battery, and
- (c) Move the arrangement rapidly towards the test coil C_{1}.
- (ii) Replace the galvanometer by a small bulb, the one of the same kind, found in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.
$$\phi_1=\pi r_1^{2}\text{B}_2\\=\frac{\mu_0\pi r_1^{2}}{2r_2}I_2=M_{12}I_{2}\\\text{Thus},\text{M}_{12}=\frac{\mu_0\pi r_1^{2}}{2r_2}\\\text{M}_{12}=\text{M}_{21}=\frac{\mu_0\pi r_1^{2}}{2r_2}$$
Q. Refer to figure the arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.
Ans. Let us first consider the forward motion from x = 0 to x = 2b. The flux fB linked with the circuit SPQR is
= Blb b ≤ x < 2b
The induced emf is,
$$\epsilon=-\frac{d\phi_nB}{dt}\\=-Blv\space\space\space0 \le x \text{\textless}\space b\\=0\space b \le x \text{\textless}\space 2b\\\text{When the induced emf is non-zero, the current I is (in magnitude)}\\\text{I}=\frac{Blv}{r}$$
$$\text{F}=\frac{B^{2}I^{2}v}{r}\space\space0\le x\text{\textless}b\\=0\space\space b\le x\text{\textless}2b\\\text{The joule heating loss is given by}\\\text{F}_{J}=\text{I}^{2}_{r}\\=\frac{B^{2}I^{2}v^{2}}{r}\space 0\le x \le b\\=0\space b\le x\text{\textless}b\\=0\space b\le x \text{\textless}2b\\$$
One obtains similar expressions for the inward motion from x = 2b to x = 0. One can appreciate the whole process by examining the sketch of various quantities displayed in figure.
Q. (i) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in figure.
(ii) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1 m and assume that the loop has a large resistance.
Ans. (i) Consider a small element dy in the loop at a distance (v) from the long straight wire.
Same has been shown in figure.
Magnetic flux associated with element dy,
dΦ = BdA
Where,
dA = Area of element dy = a dy
B = Magnetic field at distance y
$$\text{y}=\frac{\mu_0I}{2\pi y}\\\text{I = Current through wire}\\\mu_0=\text{Permeability of free space}=4\pi×10^{-7}\text{TmA}^{-1}\\\therefore\space d\phi=\frac{\mu_0Ia}{2\pi}\frac{dy}{y}\\\phi=\frac{\mu_0Ia}{2\pi}\int\frac{dy}{y}\\\text{y tends from x to a + x}\\\therefore\space\phi=\frac{\mu_0Ia}{2\pi}\int^{a+x}_{x}\frac{dy}{y}\\=\frac{\mu_0Ia}{2\pi}[\text{log}_ey]^{a+x}_{x}\\=\frac{\mu_0Ia}{2\pi}\text{log}_e\bigg(\frac{a+x}{x}\bigg)$$
$$\text{For mutual inductance M, the flux is given by}\\\phi=MI\\\therefore\space\text{MI}=\frac{\mu_0Ia}{2\pi}\text{log}_e\bigg(\frac{a+x}{x}\bigg)\\\text{M}=\frac{\mu_0Ia}{2\pi}\text{log}_e\bigg(\frac{a}{x}+1\bigg)\\\text{(ii)\space\text{Emf induced in the loop,}}\\e=B^{‘}av\\=\bigg(\frac{\mu_0I}{2\pi x}\bigg)\text{av}\\\text{Given, I = 50 A}\\\text{Velocity of loop (v) = 10 m/s}\\\text{e}=\frac{4\pi×10^{-7}×50×0.1×10}{2\pi×0.2}\\\text{e}=5×10^{-5}V$$
- (i) What is the instantaneous value of the emf induced in the wire ?
- (ii) What is the direction of the emf ?
- (iii) Which end of the wire is at the higher electrical potential ?
- Ans. Given,
- Length of the wire (l) = 10 m
- Falling speed of the wire (v) = 5.0 m/s
- Magnetic field strength (B) = 0.3 × 10^{– 4 }Wb m^{– 2}
- (i) Emf induced in the wire is given by
- e = Blv = 0.3 × 10^{– 4} × 10 × 5
- = 1.5 × 10^{– 3} V
- (ii) From Fleming’s right hand rule, it can be inferred that the induced emf is West to East.
- (iii) The eastern end of the wire is at a higher potential.
Ans. Initial flux through the coil.
Therefore, estimated value of the induced emf is,
$$\epsilon=N\frac{\Delta\phi}{\Delta t}=\frac{500×(6\pi×10^{-7})}{0.25}\\=3.8×10^{-3}V\\\text{I}=\frac{\epsilon}{R}=\frac{3.8×10^{-3}}{2}=1.9×10^{-3}A\\\text{Note that the magnitude of} \epsilon \text{and I are the estimated values}.\\\text{Their instantaneous values are different and depend upon the speed of rotation at the particular instant.}$$
Q. 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 emf developed across the cut if the velocity of the loop is 1 cm s– 1 in a direction normal to the (i) longer side, (ii) shorter side of the loop ? For how long does the induced voltage last in each case ?
Ans. Given,
Length of the rectangular wire, l = 8 cm = 0.08 m
Width of the rectangular wire, b = 2 cm = 0.02 m
Hence, area of the rectangular loop,
A = lb = 0.08 × 0.02
Magnetic field strength (B) = 0.3 T
Velocity of the loop (v) = 1 cm/s = 0.01 m/s
(i) Emf developed in the loop is given by
e = Blv
Time taken to travel along the width (t)
$$=\frac{\text{Distance}\space \text{travelled}}{\text{Velocity}}=\frac{b}{v}\\=\frac{0.02}{0.01}=2s\\\text{Hence, the induced voltage is}\space2.4 × 10^{\normalsize– 4} \text{V which lasts for 2 s.}\\\text{(ii)}\text{Emf developed (e) = Bbv}\\= 0.3 × 0.02 × 0.01\\= 0.6 × 10^{\normalsize– 4} V\\\text{Time taken to travel along the length}\\t=\frac{\text{Distance travelled}}{\text{Velocity}}=\frac{l}{v}=\frac{0.08}{0.01}=8s\\\text{Hence, the induced voltage is}\space 0.6×10^{-4}V\text{which lasts for 8 s.}$$
- Ans. Given,
- f = 0.5 Hz, N = 100, A = 0.1 m^{2} and B = 0.01 T.
- As we know that
- ε_{0} = NBA (2πv)
- = 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5
- = 0.314 V
- The maximum voltage is 0.314 V.
Ans. Given,
Speed of the jet plane (v) = 1800 km/h = 50 m/s
Wing span of jet plane (l) = 25 m
Angle of dip (δ) = 30°
Vertical component of Earth’s magnetic field is given by
Voltage difference between the ends of the wing is given by
= 3.125 V
Therefore, the voltage difference arises between the ends of the wings is 3.125 V.
Ans. Given,
Length of the solenoid (l) = 30 cm = 0.3 m
Number of turns on the solenoid (N) = 500
Current in the solenoid (I) = 2.5 A
$$\text{Average back emf, e =}\frac{d\phi}{dt}\space\text{…(i)}\\\text{Where},d\phi=\text{Change in flux}\\d\phi = \text{NAB}\space\text{…(ii)}\\\text{Where},\\\text{B}=\text{Magnetic field strength}\\\text{B}=\mu_0\frac{NI}{l}\space\text{…(iii)}\\\text{Where,}\space\mu_0=\text{Permeability of free space}\\=4\pi×10^{-7}\text{TmA}^{-1}\\\text{From equation (i), (ii) and (iii),}\\\text{e}=\frac{\mu_0N^{2}IA}{l×t}\\=\frac{4\pi× 10^{-7}×(500)^{2}×2.5×25×10^{-4}}{0.3×10^{-3}}\\ = 6.5V$$
Hence, the average back emf induced in the solenoid is 6.5 V.
Q. 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 W, 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. Given,
- Radius of the circular coil (r) = 8 cm = 0.08 m
- Area of the coil (A) = πr^{2} = π × (0.08)^{2} m^{2}
- Number of turns on the coil (N) = 20
- Angular speed (ω) = 50 rad/s
- Magnetic field strength (B) = 3 × 10^{– 2}v T
- Resistance of the loop (R) = 10 Ω
- Maximum induced emf is given by
- e = NωAB
- = 20 × 50 × π × (0.08)^{2} × 3 × 10^{– 2}
- = 0.603 V
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 by
$$\text{I}=\frac{e}{R}=\frac{0.603}{10}=0.0603\space A\\\text{Average power loss due to joule heating is given by}\\\text{P}=\frac{eI}{2}=\frac{0.603×0.0603}{2}=0.018W\\\text{The current induced in the coil produces a torque opposing the rotation of the coil.}\\\text{ The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly.}\\ \text{ Hence, dissipated power comes from the external rotor.}$$
Q. (i) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid.
(ii) How does this magnetic energy compare with the electrostatic energy stored in a capacitor ?
Ans. (i) The magnetic energy is given by
$$\text{U}_B=\frac{1}{2}LI^{2}\\=\frac{1}{2}L\bigg(\frac{B}{\mu_0n}\bigg)^{2}\\\text{(since B =} \mu_0nI,\text{for a solenoid})\\=\frac{1}{2}(\mu_0n^{2}Al)\bigg(\frac{B}{\mu_0n}\bigg)^{2}\\(as L=\mu_0 n^{2}Al)\\=\frac{1}{2\mu_0}B^{2}AL\\\text{(ii) The magnetic energy per unit volume is,}\\u_B=\frac{U_B}{V}\\(\text{where} V \text{is volume that contains flux})\\$$
$$=\frac{U_B}{Al}[\text{From part} (i) \text{of question}]\\=\frac{B^{2}}{2\mu_0}\space\text{…(i)}\\\text{We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor.}\\u_E=\frac{1}{2}\epsilon_0E^{2}\space\text{…(ii)}$$
In both the cases energy is proportional to the square of the field strength. Equations (i) and (ii) have been derived for special cases : a solenoid and a parallel plate capacitor, respectively. But they are general and valid for any region of space in which a magnetic field or/and an electric field exist.
Q. Given figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader.
Determine the direction of induced current in each loop using Lenz’s law.
- (i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field. The induced current must flow along the path bcdab so that it opposes the increasing flux.
- (ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux.
- (iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux.
- Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field.
Q. A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference
of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (Fig.).
A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring ?
Ans. As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. The magnitude of the
emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by
$$\text{d}\epsilon=Bv\space dr\\\text{Hence},\\\epsilon=\int d\epsilon=\int^{R}_{0}Bvdr\\=\int^{R}_{0}B\omega r\space dr=\frac{B\omega R^{2}}{2}\\\text{Note that we have used v =} \omega \text{r. This gives}\\\epsilon=\frac{1}{2}×1.0×2\pi×50×(1^{2})\\=157\space V$$
Q. A metallic rod CD rests on thick metallic wire PQRS with arms PQ and RS parallel to each other, at a distance l = 40 cm as shown in following figure. A uniform magnetic field B = 0.1 T acts perpendicular to the plane of this paper, pointing inwards (i.e., away from the reader). The rod is now made to slide towards right, with a constant velocity of v = 5 m/s.
- (i) How much emf is induced between the two ends of the rod CD ?
- (ii) What is the direction in which the induced current flows ?
Ans. (i) e = Bvl
= 0.1 × 5 × 0.40 = 0.2 V
(ii) Direction of induced current = D → C → Q → R (anticlockwise)