Question 1

Q. 6.9. 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?

Accepted Answer

Mutual inductance of a pair of coils, $\mu=1.5 \mathrm{H}$

Initial current, $\mathrm{I}_{1}=0 \mathrm{~A}$

Final current $\mathrm{I}_{2}=20 \mathrm{~A}$

Change in current, $d I=I_{2}-I_{1}=20-0=20 \mathrm{~A}$

Time taken for the change, $\mathrm{t}=0.5 \mathrm{~s}$

$$

\begin{equation*}

ext { Induced emf, } \quad e=\frac{d \phi}{d t} 	ag{1}

\end{equation*}

$$

Where $d \phi$ is the change in the flux linkage with the coil.

Emf is related with mutual inductance as:

$$

\begin{equation*}

e=\mu \frac{d I}{d t} 	ag{2}

\end{equation*}

$$

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

$\frac{d \phi}{d t}=\mu \frac{d I}{d t}$

$d \phi=1.5 	imes(20)$

$$

=30 \mathrm{~Wb}

$$

Hence, the change in the flux linkage is 30 Wb .

Question 2

Q. 6.1. Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).

(a)

(b)

(c)

(Tapping key just closed)

(d)

(e)

(f)

Accepted Answer

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 situations can be predicted as follows:

(a) The direction of the induced current is along qrpq.

(b) The direction of the induced current is along prqp.

(c) The direction of the induced current is along yzxy.

(d) The direction of the induced current is along zyxz.

(e) The direction of the induced current is along xryx.

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

Question 3

Q. 6.2. A 1.0 m long metallic rod is rotated with an angular frequency of $400 \mathrm{rad} \mathrm{s}^{-1}$ 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.

Accepted Answer

Length of the rod, $\mathrm{I}=1 \mathrm{~m}$

Angular frequency, $\omega=400 \mathrm{rad} / \mathrm{s}$

Magnetic field strength, $B=0.5 \mathrm{~T}$

One end of the rod has zero linear velocity, while the other end has a linear velocity of $\mathrm{I} \omega$.

Average linear velocity of the rod, $v=\frac{l \omega+0}{2}=\frac{l \omega}{2}$

Emf developed between the centre and the ring,

$$

\begin{aligned}

e & =B l v=B l\left(\frac{l \omega}{2}ight)=\frac{B l^{2} \omega}{2} \

& =\frac{0.5 	imes(1)^{2} 	imes 400}{2}=100 \mathrm{~V}

\end{aligned}

$$

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

Question 4

Q. 6.3. A long solenoid with 15 turns per cm has a small loop of area $2.0 \mathrm{~cm}^{2}$ 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?

Accepted Answer

Number of turns on the solenoid $=15$ turns/cm $=1500 \mathrm{turns} / \mathrm{m}$

Number of turns per unit length, $\mathrm{n}=1500$ turns

The solenoid has a small loop of area, $\mathrm{A}=2.0 \mathrm{~cm}^{2}=2 	imes 10^{-4} \mathrm{~m}^{2}$

Current carried by the solenoid changes from 2 A to 4 A .

$	herefore$ Change in current in the solenoid, $\mathrm{di}=4-2=2 \mathrm{~A}$

Change in time, $\mathrm{dt}=0.1 \mathrm{~s}$

Induced emf in the solenoid is given by Faraday's law as:

$$

\begin{equation*}

e=\frac{d \phi}{d t} 	ag{i}

\end{equation*}

$$

Where, $\phi=$ Induced flux through the small loop

= BA

$B=$ Magnetic field

$$

\begin{equation*}

=\mu_{0} n i 	ag{iii}

\end{equation*}

$$

$\mu_{0}=$ Permeability of free space

$=4 \pi 	imes 10^{-7} \mathrm{H} / \mathrm{m}$

Hence, equation (i) reduces to:

$$

\begin{aligned}

e & =\frac{d}{d t}(B A) \

& =A \mu_{0} n 	imes\left(\frac{d i}{d t}ight) \

& =2 	imes 10^{-4} 	imes 4 \pi 	imes 10^{-7} 	imes 1500 	imes \frac{2}{0.1} \

& =7.54 	imes 10^{-6} \mathrm{~V}

\end{aligned}

$$

Hence, the induced voltage in the loop is $7.54 	imes 10^{-6} \mathrm{~V}$.

Question 5

Q. 6.4. 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 emf developed across the cut if the velocity of the loop is $1 \mathrm{~cm} \mathrm{~s}^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in

Accepted Answer

Length of the rectangular wire, $\mathrm{I}=8 \mathrm{~cm}=0.08 \mathrm{~m}$

Width of the rectangular wire, $\mathrm{b}=2 \mathrm{~cm}=0.02 \mathrm{~m}$

Hence, area of the rectangular loop,

$\mathrm{A}=\mathrm{lb}$

$=0.08 	imes 0.02$

$=16 	imes 10^{-4} \mathrm{~m}^{2}$

Magnetic field strength, $\mathrm{B}=0.3 \mathrm{~T}$ Velocity

of the loop, $v=1 \mathrm{~cm} / \mathrm{s}=0.01 \mathrm{~m} / \mathrm{s}(\mathrm{a})$

Emf developed in the loop is given as:

$\mathrm{e}=\mathrm{Blv}$

$=0.3 	imes 0.08 	imes 0.01=2.4 	imes 10^{-4} \mathrm{~V}$

Time taken to travel along the width, $t=\frac{	ext { Distance travelled }}{	ext { Velocity }}=\frac{b}{v}$

$$

=\frac{0.02}{0.01}=2 \mathrm{~s}

$$

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

(b) Emf developed, $\mathrm{e}=\mathrm{Bbv}$

$=0.3 	imes 0.02 	imes 0.01=0.6 	imes 10^{-4} \mathrm{~V}$

Time taken to travel along the length, $t=\frac{	ext { Distance traveled }}{	ext { Velocity }}=\frac{l}{v}$

$$

=\frac{0.08}{0.01}=8 \mathrm{~s}

$$

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

Question 6

Q. 6.6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad $\mathrm{s}^{-1}$ in a uniform horizontal magnetic field of magnitude $3.0 	imes 10^{-2}$ T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance $10 \Omega$, calculate the maximum value of current in the coil. Calculate the

Accepted Answer

Max induced emf $=0.603 \mathrm{~V}$

Average induced emf $=0 \mathrm{~V}$

Max current in the coil $=0.0603 \mathrm{~A}$

Average power loss $=0.018 \mathrm{~W}$

(Power comes from the external rotor)

Radius of the circular coil, $r=8 \mathrm{~cm}=0.08 \mathrm{~m}$

Area of the coil, $\mathrm{A}=\pi \mathrm{r}^{2}=\pi 	imes(0.08)^{2} \mathrm{~m}^{2}$

Number of turns on the coil, $N=20$

Angular speed, $\omega=50 \mathrm{rad} / \mathrm{s}$

Magnetic field strength, $\mathrm{B}=3 	imes 10^{-2} \mathrm{~T}$

Resistance of the loop, $\mathrm{R}=10 \Omega$

Maximum induced emf is given as:

$\mathrm{e}=\mathrm{N} \omega \mathrm{AB}$

$=20 	imes 50 	imes \pi 	imes(0.08)^{2} 	imes 3 	imes 10^{-2}$

$=0.603 \mathrm{~V}$

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 as:

$$

\begin{aligned}

I & =\frac{e}{R} \

& =\frac{0.603}{10}=0.0603 \mathrm{~A}

\end{aligned}

$$

Average power loss due to joule heating:

$$

\begin{aligned}

P & =\frac{e I}{2} \

& =\frac{0.603 	imes 0.0603}{2}=0.018 \mathrm{~W}

\end{aligned}

$$

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.

Question 7

Q. 6.7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of $5.0 \mathrm{~m} \mathrm{~s}^{-1}$, at right angles to the horizontal component of the earth's magnetic field, $0.30 	imes 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$.

(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 i

Accepted Answer

Length of the wire, $\mathrm{I}=10 \mathrm{~m}$

Falling speed of the wire, $v=5.0 \mathrm{~m} / \mathrm{s}$

Magnetic field strength, $\mathrm{B}=0.3 	imes 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

(a) Emf induced in the wire, $\mathrm{e}=\mathrm{Blv}$

$$

\begin{aligned}

& =0.3 	imes 10^{-4} 	imes 5 	imes 10 \

& =1.5 	imes 10^{-3} \mathrm{~V}

\end{aligned}

$$

(b) Using Fleming's right hand rule, it can be inferred that the direction of the induced emf is from West to East.

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

Question 8

Q. 6.8. Current in a 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.

Accepted Answer

Initial current, $\mathrm{I}_{1}=5.0 \mathrm{~A}$

Final current, $\mathrm{I}_{2}=0.0 \mathrm{~A}$

Change in current, $d I=I_{1}-I_{2}=5 \mathrm{~A}$

Time taken for the change, $\mathrm{t}=0.1 \mathrm{~s}$

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

$$

\begin{aligned}

\mathrm{e} & =\mathrm{L} \frac{d i}{d t} \

L & =\frac{e}{\left(\frac{d i}{d t}ight)} \

& =\frac{200}{\frac{5}{0.1}}=4 \mathrm{H}

\end{aligned}

$$

Hence, the self induction of the coil is 4 H .

6. Electric Charges and Fields

Exercises - 1