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:

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 $\boldsymbol{y z x y}$.

The direction of the induced current is along $\mathbf{z y x z}$.

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.

Answer

(a) According to Lenz’s law, the magnetic flux produced by the induced current opposes the cause of induction. It defines the direction of flow of induced current.

In the given closed loop, the loop is placed in a magnetic field and it is changing its shape from irregular to circular. During this change the magnetic flux linked with it increases, so according to Lenz’s law, the induced current should produce magnetic flux such that it reduces the flux linked to the coil.

The induced magnetic flux should be in opposite direction to the original flux. Thus, the current should flow in anti clockwise direction.

Thus, the direction of the induced current is adcba.

(b) As the circular loop is being deformed into a narrow straight line, the magnetic flux linked to the loop will decrease and according to Lenz’s law, the induced current should oppose the cause of change. Therefore, the induced flux should be produced in the direction of the original flux.

Therefore, the induced current should flow in anti clockwise direction.

Thus, the direction of induced current is a ′ d ′ c ′ b ′ .

Answer

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

Number of turns per unit length, $n=1500$ turns

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

Current carried by the solenoid changes from $2 \mathrm{~A}$ to $4 \mathrm{~A}$.

$\therefore$ Change in current in the solenoid, $d i=4-2=2 \mathrm{~A}$

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

Induced $e m f$ in the solenoid is given by Faraday’s law as:

$e=\frac{d \phi}{d t}$

Where,

$\phi=$ Induced flux through the small loop

$=B A \ldots(i i)$

$B=$ Magnetic field

$=\mu_{0} n i$

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

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

Hence, equation $(i)$ reduces to:

$$ \begin{aligned} e & =\frac{d}{d t}(B A) \\ & =A \mu_{0} n \times\left(\frac{d i}{d t}\right) \\ & =2 \times 10^{-4} \times 4 \pi \times 10^{-7} \times 1500 \times \frac{2}{0.1} \\ & =7.54 \times 10^{-6} \mathrm{~V} \end{aligned} $$

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

Answer

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

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

Hence, area of the rectangular loop,

$A=l b$

$=0.08 \times 0.02$

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

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

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

Emf developed in the loop is given as:

$e=B l v$

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

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

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

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

Emf developed, $e=B b v$

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

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

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

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

Answer

$$ 1 = 1.0 \mathrm{~cm} \quad \omega=400 \mathrm{rad} / \mathrm{s} $$

$\mathrm{B}=0.5 \mathrm{~T}$

$$ \begin{aligned} \varepsilon= & -\frac{\mathrm{d} \Phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B} \cdot \frac{\pi \mathrm{r}^{2} \theta}{2 \pi}\right)=\mathrm{B}\left(\frac{1}{2} \mathrm{r}^{2} \omega\right) \\ & =100 \mathrm{~V} \end{aligned} $$

Answer

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

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

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

Emf induced in the wire,

$$ \begin{aligned} e & =B l v \\ & =0.3 \times 10^{-4} \times 5 \times 10 \\ & =1.5 \times 10^{-3} \mathrm{~V} \end{aligned} $$

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

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

Answer

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

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

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

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

Average emf, $e=200 \mathrm{~V}$

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

$$ \begin{aligned} e & =L \frac{d i}{d t} \\ L & =\frac{e}{\left(\frac{d i}{d t}\right)} \\ & =\frac{200}{\frac{5}{0.1}}=4 \mathrm{H} \end{aligned} $$

Hence, the self induction of the coil is $4 \mathrm{H}$.

Answer

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

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

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

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

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

Induced emf, $e=\frac{d \phi}{d t}$

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

Emf is related with mutual inductance as:

$$ \begin{equation*} e=\mu \frac{d I}{d t} \tag{2} \end{equation*} $$

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

$$ \begin{aligned} \frac{d \phi}{d t} & =\mu \frac{d I}{d t} \\ d \phi & =1.5 \times(20) \\ & =30 \mathrm{~Wb} \end{aligned} $$

Hence, the change in the flux linkage is $30 \mathrm{~Wb}$.