Class 12 Physics Chapter 6 NCERT Solutions: Free PDF, Important Formulas and Questions

6.1

The direction of the induced current in a closed loop is given by Lenz's law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Using Lenz's rule, the direction of the induced current in the given situation can be predicted as follows:

The direction of the induced current is along 'qrpq'.

The direction of the induced current is along 'prqp'.

The direction of the induced current is along 'yzxy'.

The direction of the induced current is along 'zyxz'.

The direction of the induced current is along 'xryx'.

No current is induced since the field lines are lying in the plane of the closed loop.

6.2 (a) As the loop changes from irregular to circular shape, its area increases. Hence the magnetic flux linked with it also increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of the original flux. For this induced current should flow in the anti-clockwise direction, i.e. adcb.

(b) As this circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic flux linked also decreases. By Lenz's law, the induced current should produce a flux in the direction of the original flux. For this the induced current should flow in the anti-clock wise direction, i.e. a'd'c'b'.

6.3 Number of turns on the solenoid = 15 turns per cm = 1500 turns per m

Hence, number of turns per unit length, n = 1500

The loop area in the solenoid, A = 2.0 ${cm}^2$ = 2 $\times {10}^{-4}$ $m^2$

Current carrying by the solenoid, changes from 2 .0 A to 4.0 A

So, change of current, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

According to Faraday’s law, the induced emf, e = $\frac{d\phi }{dt}$ …………. (i)

Where $\phi$ = induced flux through the small loop = BA …………. (ii)

Magnetic field is given by B = ${\mu }_{0}ni$ ……………………………. (iii)

Where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ H/m

From equation (i),

e = $\frac{d}{dt}\left(BA\right)$

= A ${\mu }_{0}n\times \frac{di}{dt}$

= 2 $\times {10}^{-4}\times$ 4 $\pi \times {10}^{-7}\times 1500\times \frac{2}{0.1}$

= 7.54 $\times {10}^{-6}$ V

Hence the induced emf while current is changing is 7.54 $\times {10}^{-6}$ V

6.4 Length of the rectangular wire loop, l = 8 cm = 0.08 m

Width of the rectangular loop, b = 2 cm = 0.02 m

Hence, the area of the rectangular loop, A = l $\times b$ = 0.08 $\times 0.02$ = 1.6 $\times {10}^{-3}$ $m^2$

Magnetic field strength, B= 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m /s

Emf developed in the longer side:

Emf developed in the loop is given by e = Blv = 0.3 $\times 0.08\times 0.01$ = 2.4 $\times {10}^{-4}$ V

Time taken to travel the width = $\frac{Distancetravelled}{velocity}$ = $\frac{b}{v}$ = $\frac{0.02}{0.01}$ s= 2 s

Hence the induced voltage is 2.4 $\times {10}^{-4}$ V, which lasts for 2 s.

Emf developed in the shorter side:

Emf developed in the loop is given by e = Bbv = 0.3 $\times 0.02\times 0.01$ = 0.6 $\times {10}^{-4}$ V

Time taken to travel the width = $\frac{Distancetravelled}{velocity}$ = $\frac{l}{v}$ = $\frac{0.08}{0.01}$ s= 8 s

Hence the induced voltage is 0.6 $\times {10}^{-4}$ V, which lasts for 8 s.

6.5 Length of the rod, l = 1.0 m

Angular frequency,  $\omega$ = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of l $\omega$

The average linear velocity, v = $\frac{0+l\omega }{2}$ = $\frac{l\omega }{2}$

emf developed between the centre and the ring is given by

e = Blv = Bl ( $\frac{l\omega }{2})$ = $\frac{B{l}^2\omega }{2}$ = $\frac{0.5\times 1^2\times 400}{2}$ = 100 V

Therefore, the emf developed between the centre and the ring is 100 V.

6.6 Given,

Radius of the circular coil, r = 8.0 cm = 0.08 m

Area of the coil, A = $\pi r^2$ = $\pi \times {0.08}^2$ = 20.106 $\times {10}^{-3}$ $m^2$

Number of turns in the coil, n = 20

Angular speed, $\omega$ = 50 rad/s

Magnetic field strength, B = 3.0 $\times {10}^{-2}$ T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as e = n $\omega AB$ = 20 $\times 50\times$ 20.106 $\times {10}^{-3}\times$ 3.0 $\times {10}^{-2}$ = 0.603 V

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given by, $I_{max}$ = $\frac{e}{R}$ = $\frac{0.603}{10}$ = 0.0603 A

Average power loss due to Joule heating is given by, $P_{loss}$ = $\frac{e{I}_{max}}{2}$ = 0.018 W

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent; it must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

6.7 Length of the wire, l = 10 m

Falling speed of the wire, v = 5 m/s

Magnetic field strength, B = 0.30 $*{10}^{-4}$ Wb/ $m^2$

The instantaneous emf induced in the wire, e = Blv = 0.30 $*{10}^{-4}*10*$ 5 = 1.5 $*{10}^{-3}$ V

Using Fleming's right hand rule, the direction of the induced emf is from West to East.

The eastern end of the wire is at a higher potential.

6.8 Given:

Initial current,  $I_1$ = 5.0 A

Final current,  $I_2$ = 0 A

Change in current, di = $I_1-I_2$ = 5 A

Time taken for the change, dt = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, L is given by

L = $\frac{e}{\frac{di}{dt}}$ = $\frac{200}{\frac{5}{0.1}}$ = 4 H

Hence, the self-inductance of the coil is 4 H.

6.9 Mutual inductance of a pair of coils, $\mu$ = 1.5 H

Initial current, $I_1$ = 0 A

Final current, $I_2$ = 20 A

Change in current, dI = $I_2$ - $I_1$ = 20 A

Time taken for change, dt = 0.5 s

From the relation of induced emf e = $\frac{d\varnothing }{dt}$ , where $d\varnothing$ is the change in the flux linkage with the coil ……………(1)

The relation between emf and mutual inductance is e = $\mu \frac{dI}{dt}$ ………….. (2)

Combining equations (1) and (2), we get

$\frac{d\varnothing }{dt}=\mu \frac{dI}{dt}$ or $d\varnothing =\mu dI$ = 1.5 $\times 20$ = 30 Wb

Hence, the change in the flux linkage is 30 Wb

6.11 The area of the rectangular coil, A = 8 $\times 2=$ 16 ${cm}^2$ = 16 $\times {10}^{-4}$ $m^2$

Initial value of the magnetic field, $B_1$ = 0.3 T

Rate of decrease of the magnetic field, $\frac{dB}{dt}$ = 0.02 T/s

From the relation of induced emf e = $\frac{d\varnothing }{dt}$ , where $d\varnothing$ is the change in the flux linkage with the coil = A $\times B$

Hence, e = $\frac{d\left(AB\right)}{dt}$ = A $\frac{dB}{dt}$ = 16 $\times {10}^{-4}\times 0.02$ = 3.2 $\times {10}^{-5}$ V

Resistance in the loop, R = 1.6 Ω

Hence I = $\frac{e}{R}$ = $\frac{3.2\mathrm{}\times {10}^{-5}}{1.6}$ = 2 $\times {10}^{-5}$ A

Power dissipated in the form of heat is given by

P = $I^2$ R = ( ${\mathrm{}2\mathrm{}\times {10}^{-5})}^2\times 1.6$ = 6.4 $\times {10}^{-10}$ W

The source of heat loss is an external agent, which is responsible for changing the magnetic field with time.

6.12 Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 $\times 0.12=0.0144m^2$

Velocity of the loop, v = 8 cm /s = 0.08 m/s

Gradient of the magnetic field along negative x – direction.

$\frac{dB}{dx}$ = ${10}^{-3}$ T/cm = ${10}^{-1}$ T/m

Rate of decrease of magnetic field

$\frac{dB}{dt}$ = ${10}^{-3}$ T/s

Resistance of the loop, R = 4.50 mΩ = 4.5 $\times {10}^{-3}$ Ω

Rate of change of the magnetic flux due to motion of the loop in a non-uniform magnetic field is given as:

$\frac{d\varnothing }{dt}$ = A $\times \frac{dB}{dx}$ $\times v$ = $0.0144\times {10}^{-1}\times 0.08$ = 1.152 $\times {10}^{-4}$ T $m^2{s}^{-1}$

Rate of change of flux due to explicit time variation in field B is given as:

$\frac{d\varnothing \text{'}}{dt}$ = A $\times \frac{dB}{dt}$ = $0.0144\times {10}^{-3}$ = 1.44 $\times {10}^{-5}$ T $m^2{s}^{-1}$

Since the rate of change of the flux is the induced emf, the total emf in the loop can be calculated as:

e = 1.152 $\times {10}^{-4}$ + 1.44 $\times {10}^{-5}$ = 1.296 $\times {10}^{-4}$ V

The induced current, I = $\frac{e}{R}$ = $\frac{1.296\mathrm{}\times {10}^{-4}}{4.5\mathrm{}\times {10}^{-3}}$ = 28.8 $\times {10}^{-3}$ A

Therefore, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

6.13 Area of the coil, A = 2 ${cm}^2$ = 2 $\times {10}^{-4}$ $m^2$

Number of turns, n = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 $\times {10}^{-3}$ C

Total resistance of the coil and galvanometer, R = 0.50 Ω

We know, Induced current in the coil, I = $\frac{Inducedemf\left(e\right)}{R}$ ……… (1)

Induced emf is given by the equation, e = $-$ n $\frac{d\varnothing }{dt}$ ………………… (2),

Where d $\varnothing$ is the change of flux

Combining equations (1) and (2), we get

I = $\frac{-\mathrm{n}\frac{d\varnothing }{dt}}{R}$ or Idt = $-\frac{n}{R}$ d $\varnothing$ ……… (3)

Initial flux through coil, ${\varnothing }_i$ = BA, where B = Magnetic field strength

Final flux through coil, ${\varnothing }_f$ = 0

Integrating equation (3), we get

$\int Idt$ = $-\frac{n}{R}{\int }_{{\varnothing }_i}^{{\varnothing }_f}d\varnothing$

$\int Idt=Q=\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{f}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}$

Therefore

Q = $-\frac{n}{R}\left({\varnothing }_f-{\varnothing }_i\right)=-\frac{n}{R}\left(0-{\varnothing }_i\right)=\frac{n{\varnothing }_i}{R}=\frac{nBA}{R}$

B = $\frac{QR}{nA}$ = $\frac{7.5\mathrm{}\times {10}^{-3}\times 0.5}{25\times 2\times {10}^{-4}}$ = 0.75 T

Hence, the field strength is 0.75 T

6.14 Length of the rod, l = 15 cm = 0.15 m

Magnetic field strength, B = 0.5 T

Resistance of the closed loop, R = 9 mΩ = 9 $\times {10}^{-3}$ Ω

Induced emf = 9 mV

Polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

Speed of the rod, v = 12 cm/s = 0.12 m/s

Induced emf is given as, e = Bvl = 0.5 $\times 0.12\times 0.15$ = 9 mV

Yes, when key K is closed, excess charge is maintained by the continuous flow of current. When the key K is open, there is excess charge built up at both ends of the rods.

Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod. There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

Retarding force exerted on the rod, F = IBl , where

I = Current flowing through the rod = $\frac{e}{R}$ = $\frac{9\times {10}^{-3}}{9\mathrm{}\times {10}^{-3}}$ = 1 A

F = 1 $\times 0.5\times 0.15$ = 0.075 N

Speed of the rod, v = 12 cm/s = 0.12 m/s

Power is given by P = Fv = 0.075 $\times 0.12=9\times {10}^{-3}$ W = 9 mW

When key K is open, no power is expended.

Power dissipated by heat = $I^{2}R=1\times$ 9 $\times {10}^{-3}$ W = 9 mW

The source of this power is an external agent.

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.

6.15 Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 ${cm}^2$ = 25 $\times {10}^{-4}$ $m^2$

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flowing time, t = ${10}^{-3}$ s

We know average back emf, e = $\frac{d\varnothing }{dt}$ …………………. (1)

Where $d\varnothing =$ Change in flux = NAB ……………… .(2)

B = Magnetic field strength = ${\mu }_0\frac{NI}{l}$ ………………... (3)

Where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

From equation (1) and (2), we get

e = $\frac{NAB}{dt}$ = $\frac{NA}{dt}\times {\mu }_0\frac{NI}{l}$ = $\frac{{\mu }_0{N}^{2}AI}{lt}$ = $\frac{4\pi \times {10}^{-7}\times {500}^2\times 25\mathrm{}\times {10}^{-4}\times 2.5}{0.3\times {10}^{-3}}$ = 6.544 V

Hence the average back emf induced in the solenoid is 6.5 V

6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.

Magnetic flux associated with element dy, d $\varnothing$ = BdA, where

dA = Area of the element dy = a dy

Magnetic flux at distance y, B = $\frac{{\mu }_{0}I}{2\pi y}$ , where

I = current in the wire

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Therefore,

d $\varnothing$ = $\frac{{\mu }_{0}I}{2\pi y}$ a dy = $\frac{{\mu }_{0}Ia}{2\pi }$ $\frac{dy}{y}$

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }\int \frac{dy}{y}$

Now from the figure, the range of y is x to x+a. Hence,

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }{\int }_x^{x+a}\frac{dy}{y}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\left[{\log }_e?y\right]}_x^{a+x}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

For mutual inductance M, the flux is given as

$\varnothing =MI$ . Hence

$MI=\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

M = $\frac{{\mu }_{0}a}{2\pi }{\log }_e?(\frac{a}{x}+1)$

Emf induced in the loop, e = Bav

= ( $\frac{{\mu }_{0}I}{2\pi x}$ ) $\times$ av

For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

e = $\frac{4\pi \times {10}^{-7}\times 50\times 0.1\times 10}{2\pi \times 0.2}$ = 5 $\times {10}^{-5}$ V

6.17 Line charge per unit length = $\lambda$ = $\frac{Totalcharge}{Length}$ = $\frac{Q}{2\pi r}$

Where r = distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, $\stackrel{?}{B}$ = $-B_0\stackrel{?}{k}$

At a distance r, the magnetic force is balanced by the centripetal force. i.e.

BQ $v$ = $\frac{M{v}^2}{r}$ , where v = linear velocity of the wheel. Then,

B $\times 2\lambda \pi r$ = $\frac{Mv}{r}$

v = $\frac{2B\lambda \pi r^2}{M}$

Angular velocity, $\omega =\frac{v}{R}$ = $\frac{2B\lambda \pi r^2}{MR}$

For r $\le a\le R,weget$ $\omega =-\frac{2B_0\lambda \pi a^2}{MR}\stackrel{?}{k}$

The process of creating the electric current in a conductor by moving the conductor through the magnetic field or by changing the magnetic field around the conductor is known as the Electromagnetic Induction.

The basic principle of the electromagnet is the electromagnetism. When the ferromagnetic iron core is coiled by a wire and a current is passed through the core, it produces a strong and magnetic field, and turns the setup into a magnet. By changing the current, one can change the strength of the magnetism, and when the current is switched off, the magnetism disappears.

As per the Faraday's Law of Induction, when a conductor is exposed to a changing magnetic field, a circulating electric current is induced, it is called the Eddy current. The Eddy currents have many applications including - non-destructive testing and induction furnaces.

Ohm's Law states that when the temperature remains constant, the current (I) flowing through a conductor is directly proportional to voltage (v), and inversely proportional to resistance (R).

The sum of a magnetic field that passes through a given surface area is called the magnetic flux. The mathematic formula is? = BA cos (? ).