Electromagnetic Induction

$ \begin{aligned} &\text {Self inductance }\\ &\begin{aligned} & L=\frac{\mu_0 N^2 A}{l} \\ & =\frac{4 \pi \times 10^{-7} \times(200)^2 \times\left(\frac{7}{2} \times 10^{-2}\right)^2 \times 2 \pi}{0.4} \\ & =484 \times 10^{-6} \mathrm{H}=484 \mu \mathrm{H} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Induced emf, }\\ &\begin{aligned} e & =N \frac{d \phi}{d t}=N \frac{d}{d t} B A=N A \frac{d B}{d t} \\ & =100\left(10 \times 10 \times 10^{-4}\right)(0.7) \\ & =0.7 \mathrm{volt} \end{aligned} \end{aligned} $

Let's restate the given data:

$ \begin{aligned} \text{Initial current, } I_1 &= 2 \, \text{A} \\ \text{Final current, } I_2 &= 5 \, \text{A} \\ \text{Change in current, } \Delta I &= 5 - 2 = 3 \, \text{A} \\ \text{Time, } \Delta t &= 0.3 \, \text{s} \\ \text{Induced emf, } e &= 40 \, \text{mV} = 40 \times 10^{-3} \, \text{V} \end{aligned} $

Formula:

$ e = L \frac{\Delta I}{\Delta t} $

Rearranging for $L$:

$ L = \frac{e \, \Delta t}{\Delta I} $

Substituting values:

$ L = \frac{40 \times 10^{-3} \times 0.3}{3} $

$ L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H} $

$ L = 4 \, \text{mH} $

✅ Final Answer:

Option B — 4 mH

We need to find the current in the coil at $t = 2$ seconds.

Step 1: Find the induced EMF.

The magnetic flux is given by $\phi = 50 t^2 + 4$.

To find the EMF ($E$), we take the derivative of $\phi$ with respect to $t$:

$ E = \frac{d\phi}{dt} $

Now, differentiate $\phi$:

$\frac{d}{dt}(50 t^2 + 4) = 100 t$ (because the derivative of $t^2$ is $2t$, and 50 times $2t$ is $100t$; the derivative of a constant is $0$).

Step 2: Calculate EMF at $t = 2$ seconds.

Plug in $t = 2$:

$E = 100 \times 2 = 200$ volts

Step 3: Find the current using Ohm’s Law.

Ohm’s Law: $I = \frac{E}{R}$

Here, $E = 200$ volts and $R = 200\,\Omega$.

$I = \frac{200}{200} = 1$ A

So, the current induced in the coil at $t = 2$ seconds is $1$ ampere.

We are given: current $I = 4~\mathrm{mA} = 4 \times 10^{-3}~\mathrm{A}$ and flux $\phi = 32 \times 10^{-6}~\mathrm{Tm}^2$.

Step 1: Find the inductance ($L$).

The relationship between flux, inductance, and current is $\phi = L I$.

So, $L = \frac{\phi}{I} = \frac{32 \times 10^{-6}}{4 \times 10^{-3}} = 8 \times 10^{-3}~\mathrm{H}$.

Step 2: Find the energy stored in the inductor.

The formula for energy stored is $E = \frac{1}{2} L I^2$.

Plug in the values: $E = \frac{1}{2} \times 8 \times 10^{-3} \times (4 \times 10^{-3})^2$.

Calculate the value: $E = 64 \times 10^{-9}~\mathrm{J}$.

The induced emf in a rotating spoke

$ \begin{aligned} e & =\frac{1}{2} B \omega r^2 \\ \therefore \quad e_1 & =\frac{1}{2} B\left(2 \pi \times \frac{180}{60}\right)(0.4)^2=0.48 \pi B \\ \text { and } e_2 & =\frac{1}{2} B\left(2 \pi \times \frac{90}{60}\right) \times(0.4)^2 \\ & =0.24 \pi B \\ & =2 \times 0.24 \pi \times \frac{B}{2}=\frac{e_1}{2}=\frac{E}{2}=0.5 B \end{aligned} $

$ \begin{aligned} & \text { Induced emf, } e=\frac{1}{2} B \omega R^2 \\ & =\frac{1}{2} \times 5 \times 10^{-2} \times 60 \times(0.3)^2 \\ & =0.135 \text { volt } \end{aligned} $

Given:

• Number of turns, $ N = 45 $

• Radius, $ r = 4~\text{cm} = 0.04~\text{m} $

• Change in magnetic field, $ \Delta B = 0.70~\text{T} - 0 = 0.70~\text{T} $

• Time interval, $ \Delta t = 220~\text{s} $

• The plane of the coil is perpendicular to the magnetic field, so flux is maximum.

Step 1: Rate of change of magnetic field

$ \frac{dB}{dt} = \frac{0.70}{220} = 3.18 \times 10^{-3}~\text{T/s} $

Step 2: Area of the coil

$ A = \pi r^2 = \pi (0.04)^2 = \pi \times 1.6 \times 10^{-3} = 5.027 \times 10^{-3}~\text{m}^2 $

Step 3: Induced emf (Faraday’s law)

$ \varepsilon = N A \frac{dB}{dt} $

Substitute values:

$ \varepsilon = 45 \times 5.03 \times 10^{-3} \times 3.18 \times 10^{-3} $

$ \varepsilon = 45 \times 1.6 \times 10^{-5} = 7.2 \times 10^{-4}~\text{V} $

$ \varepsilon = 0.72~\text{mV} $

✅ Final Answer:

$ 0.72~\text{mV} $

Correct option: C

$ \begin{aligned} &\text { Velocity of wire }\\ &\begin{aligned} & \quad v^2=u^2+2 g h \\ & \quad=0+2 \times 10 \times 20=400 \\ & \therefore \quad v=20 \mathrm{~m} / \mathrm{s} \\ & \therefore \text { Induced current }=\frac{e}{R}=\frac{B v l}{R} \\ & \quad=\frac{2 \times 10^{-5} \times 20 \times 30}{40}=3 \times 10^{-4} \mathrm{~A} \\ & \quad=0.3 \times 10^{-3} \mathrm{~A}=0.3 \mathrm{~mA} \end{aligned} \end{aligned} $

Given:

Initial current, $ I_1 = 14 \, \text{A} $

Final current, $ I_2 = 4 \, \text{A} $

Change in current, $ \Delta I = I_2 - I_1 = 4 \, \text{A} - 14 \, \text{A} = -10 \, \text{A} $

Time duration, $ \Delta t = 0.2 \, \text{ms} = 0.2 \times 10^{-3} \, \text{s} $

Average induced emf, $ \varepsilon = 150 \, \text{V} $

To find the self-inductance $ L $ of the circuit, we use the formula for induced emf in terms of self-inductance:

$ \varepsilon = L \left(\frac{dI}{dt}\right) $

Rearrange to solve for $ L $:

$ L = \frac{\varepsilon}{\left(\frac{dI}{dt}\right)} $

The rate of change of current $ \left(\frac{dI}{dt}\right) $ can be calculated as:

$ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{-10 \, \text{A}}{0.2 \times 10^{-3} \, \text{s}} $

Plug the values into the formula:

$ L = \frac{150 \, \text{V}}{\frac{-10 \, \text{A}}{0.2 \times 10^{-3} \, \text{s}}} $

Calculate:

$ L = \frac{150 \times 0.2 \times 10^{-3}}{-10} $

$ L = 3 \, \text{mH} $

Thus, the self-inductance of the circuit is $ 3 \, \text{mH} $.

$ \begin{aligned} &\text { Given, } l=0.1 \mathrm{~m}\\ &\begin{aligned} & \text { Resistance of wire }=1 \Omega \\ & \qquad \begin{array}{l} B=2 \mathrm{~Wb} / \mathrm{m}^2 \\ i=1 \mathrm{~mA}=10^{-3} \mathrm{~A} \end{array} \end{aligned} \end{aligned} $

Calculate equivalent resistance for Wheatstone bridge,

$ \begin{aligned} & R_1=3+3=6 \Omega \\ & R_2=3+3=6 \Omega \\ & R_{\mathrm{eq}}=\frac{R_1 \times R_2}{R_1+R_2}=\frac{6 \times 6}{6+6}=\frac{36}{12}=3 \Omega \end{aligned} $

So, total resistance in the circuit is,

$ R_{\text {total }}=3+1=4 \Omega $

The induced emf in the loop,

$ e=B v l $

So, induced current, $i=\frac{e}{R}=\frac{B v l}{R} \Rightarrow v=\frac{i R}{B l}$

$ \begin{aligned} & v=\frac{10^{-3} \times 4}{2 \times 0.1} \\ & v=2 \times 10^{-2} \mathrm{~m} / \mathrm{s}=2 \mathrm{~cm} / \mathrm{s} \end{aligned} $

Given that a coil with inductance $L$ is divided into six equal parts, these parts are connected in parallel. To find the resultant inductance of this parallel combination, we use the formula for inductors in parallel:

$ \frac{1}{L_{\text{parallel}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + \frac{1}{L_4} + \frac{1}{L_5} + \frac{1}{L_6} $

Since each part has equal inductance, we can express each as:

$ L_1 = L_2 = L_3 = L_4 = L_5 = L_6 = \frac{L}{6} $

Substituting into the parallel inductance formula gives:

$ \frac{1}{L_{\text{parallel}}} = 6 \times \frac{1}{L / 6} $

Simplifying this, we find:

$ L_{\text{parallel}} = \frac{L}{36} $

Therefore, the resultant inductance of the combination is $\frac{L}{36}$.

Given the following parameters:

Number of turns, $ N = 120 $

Inductance, $ L = 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H} $

Current, $ I = 30 \, \text{mA} = 30 \times 10^{-3} \, \text{A} $

The magnetic flux linked with the coil can be calculated using the formula:

$ \phi = \frac{L \cdot I}{N} $

Substituting the given values into the formula:

$ \phi = \frac{40 \times 10^{-3} \times 30 \times 10^{-3}}{120} $

Simplifying this:

$ \phi = 10 \times 10^{-6} \, \text{Wb} $

Thus, the magnetic flux linked with the coil is $ 10 \times 10^{-6} \, \text{Wb} $.

Given:

Induced current in circuit $Y$, $i_Y = 60 \times 10^{-4} \, \mathrm{A}$

Resistance of circuit $Y$, $R_Y = 4 \, \Omega$

Mutual inductance, $m = 3 \, \mathrm{mH} = 3 \times 10^{-3} \, \mathrm{H}$

Time, $t = 0.02 \, \mathrm{s}$

Formula:

The induced EMF ($\varepsilon_Y$) in circuit $Y$ is given by:

$ \left|\varepsilon_Y\right| = m \frac{d i_X}{d t} $

The current in circuit $Y$ due to the induced EMF can be given by:

$ i_Y R_Y = m \frac{d i_X}{d t} $

Calculation:

Substitute the known values into the equation:

$ \begin{aligned} d i_X &= \frac{60 \times 10^{-4} \times 4 \times 0.02}{3 \times 10^{-3}} \\ &= \frac{60 \times 10^{-4} \times 8 \times 10^{-2}}{3 \times 10^{-3}} \\ &= 160 \times 10^{-6} \times 10^{3} \\ &= 160 \times 10^{-3} \\ &= 0.16 \, \mathrm{A} \end{aligned} $

Thus, the change in current required in circuit $X$ is $0.16 \, \mathrm{A}$.

For solenoid $ A $:

Let $ n $ be the number of turns per unit length for solenoid $ A $, and $ l $ be its length.

The total number of turns $ N_A $ is given by:

$ N_A = n \cdot l $

For solenoid $ B $:

The number of turns per unit length for solenoid $ B $ is $ 3n $ (since the ratio is $ 1:3 $).

The length of solenoid $ B $ is $ 2l $.

The total number of turns $ N_B $ is:

$ N_B = (3n) \times (2l) = 6nl $

Self-Inductance Formulas:

The self-inductance $ L $ of a solenoid is given by the formula:

$ L = \frac{\mu_0 N^2 A}{l} $

where $ \mu_0 $ is the permeability of free space, $ N $ is the number of turns, $ A $ is the cross-sectional area, and $ l $ is the length.

For solenoid $ A $:

$ L_A = \frac{\mu_0 (N_A)^2 A}{l} = \frac{\mu_0 (n \cdot l)^2 A}{l} = \mu_0 n^2 l A $

For solenoid $ B $:

$ L_B = \frac{\mu_0 (N_B)^2 A}{l} = \frac{\mu_0 (6n \cdot l)^2 A}{2l} = \frac{\mu_0 36n^2 l^2 A}{2l} = 18 \mu_0 n^2 l A $

Ratio of Self-Inductances:

Now, the ratio of the self-inductances $ \frac{L_A}{L_B} $ is:

$ \frac{L_A}{L_B} = \frac{\mu_0 n^2 l A}{18 \mu_0 n^2 l A} = \frac{1}{18} $

Thus, the ratio of the self-inductances of solenoids $ A $ and $ B $ is $ 1:18 $.

To find the magnitude of the induced electric field at a distance $ r $ from the $ Z $-axis in the presence of a time-dependent magnetic field, we begin with the magnetic flux:

Magnetic Flux

Given that the magnetic field $ \mathbf{B} $ is expressed as $ B = B_0 r t \hat{\mathbf{k}} $, we can calculate the magnetic flux $ \phi $ through a circular region of radius $ r $:

$ \phi = \int \mathbf{B} \cdot d\mathbf{A} = \int B \cdot 2 \pi r \, dr $

Substituting the expression for $ B $:

$ \phi = 2 \pi B_0 t \int r^2 \, dr = 2 \pi B_0 t \frac{r^3}{3} $

Induced Electric Field

According to Faraday’s law of electromagnetic induction, the induced electromotive force (EMF) $\oint \mathbf{E} \cdot d\mathbf{l}$ is related to the rate of change of magnetic flux:

$ \oint \mathbf{E} \cdot d\mathbf{l} = \frac{d\phi}{dt} $

Substituting the expression for $\phi$:

$ E(2 \pi r) = \frac{d}{dt}\left(2 \pi B_0 t \frac{r^3}{3}\right) $

Differentiating with respect to time $ t $:

$ E(2 \pi r) = \frac{2 \pi B_0 r^3}{3} $

Solving for $ E $:

$ E = \frac{2 \pi B_0 r^3}{3 \times 2 \pi r} = \frac{B_0 r^2}{3} $

Therefore, the magnitude of the induced electric field at a distance $ r $ from the $ Z $-axis is given by:

$ E = \frac{B_0 r^2}{3} $

The magnetic energy stored in an inductor is given by the formula:

$ U = \frac{1}{2} L I^2 $

Where:

$ U $ is the energy stored,

$ L $ is the inductance,

$ I $ is the current through the inductor.

In this case, the current increases from $ I_1 = 2 \, \mathrm{A} $ to $ I_2 = 3 \, \mathrm{A} $.

Calculating the initial energy stored $ U_1 $ when the current is $ I_1 $:

$ U_1 = \frac{1}{2} L I_1^2 = \frac{1}{2} \times L \times 4 = 2L $

Calculating the energy stored $ U_2 $ when the current is $ I_2 $:

$ U_2 = \frac{1}{2} L I_2^2 = \frac{1}{2} \times L \times 9 = 4.5L $

The increase in energy stored, $\Delta U$, is:

$ \Delta U = U_2 - U_1 = 4.5L - 2L = 2.5L $

To find the percentage increase in energy stored:

$ \text{Percentage Increase} = \frac{\Delta U}{U_1} \times 100\% = \frac{2.5L}{2L} \times 100\% = 125\% $

Given :

Radius of the circular loop of wire :

$ r = 14 \, \mathrm{cm} = 0.14 \, \mathrm{m} $

Rate of change of the magnetic field with respect to time :

$ \frac{dB}{dt} = 0.05 \, \mathrm{T s}^{-1} $

Using Faraday's law of electromagnetic induction, the induced emf $ |e| $ is given by :

$ |e| = \left| \frac{d\Phi}{dt} \right| $

Where the magnetic flux $ \Phi $ through the loop is :

$ \Phi = B \cdot A $

Here, $ A $ is the area of the loop. Therefore :

$ \frac{d\Phi}{dt} = A \cdot \frac{dB}{dt} = \pi r^2 \cdot \frac{dB}{dt} $

Substituting the given values :

$ \begin{aligned} |e| &= \pi \left(0.14\right)^2 \cdot 0.05 \\\\ &= \frac{22}{7} \times 0.14 \times 0.14 \times 0.05 \\\\ &= 0.00308 \, \mathrm{V} \\\\ &= 3.08 \times 10^{-3} \, \mathrm{V} \\\\ &= 3.08 \, \mathrm{mV} \end{aligned} $

No. of turns in circular coil, $N=100$

Radius, $r=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$

Resistance, $R=4 \Omega$

No. of turns per unit length in the solenoid, $n=200$ turns/$\mathrm{cm}=20000$ turns/$\mathrm{m}$

$\because$ Magnetic field inside the solenoid, $B=\mu_0 n I$

Hence, induced emf

$\begin{aligned} e & =\frac{N d}{d t} B A \\ & =\frac{N d}{d t} \mu_0 n I A=N \mu_0 n A \frac{d I}{d t} \\ & =100 \times 4 \pi \times 10^{-7} \times 20000 \times \pi r^2\left(\frac{2-0}{0.04}\right) \\ & =4 \pi \times 10^{-7} \times 2 \times 10^6 \times \pi \times\left(2 \times 10^{-2}\right)^2 \times 50 \end{aligned}$

$\begin{aligned} & \text { Induced current, } I=\frac{e}{R} \\ & \qquad=\frac{4 \pi \times 10^{-7} \times 2 \times 10^6 \times \pi \times\left(2 \times 10^{-2}\right)^2 \times 50}{4} \\ & =4 \pi^2 \mathrm{~mA} . \end{aligned}$

Given, number of turns, $n=100$

Cross-sectional area, $A=3 \mathrm{~m}^2$

Angular speed, $\omega=60 \mathrm{rads}^{-1}$

Magnetic field, $B=0.04 \mathrm{~T}$

Resistance, $R=360 \Omega$

Let, maximum power be $P$,

Internal resistance, $r=R$

and equivalent resistance $=R_{\text {eq }}$

Since, $\operatorname{emf}(\varepsilon)=B n A \omega$

and $P=\varepsilon^2 / R_{\text {eq }}$

$\begin{aligned} \therefore \quad P & =\frac{B^2 n^2 A^2 \omega^2}{2 R}\left(\because \text { For max. } P, R_{\mathrm{eq}}=r+R\right) \\ & =\frac{(0.04 \times 100 \times 3 \times 60)^2}{2 \times 360} \\ & =\frac{720 \times 720}{360 \times 2}=720 \mathrm{~W} \end{aligned}$

Magnetic flux ($\phi$) is given by the formula:

$ \phi = \mathbf{B} \cdot \mathbf{A} \\ = B A \cos \theta $

Here, $B$ is the magnetic field, and $A$ is the area.

The dot product ($\cdot$) in the formula means that magnetic flux is a scalar quantity, not a vector.

This means magnetic flux does not have a direction, only a size (magnitude).

The value of $\phi$ depends on the angle $\theta$ between the field and the area:

• If $\theta = 0^\circ$, then $\phi = B A$, which is positive.
• If $\theta = 90^\circ$, then $\phi = 0$.
• If $\theta = 180^\circ$, then $\phi = -B A$, which is negative.

So, the Assertion (A) is false because magnetic flux is not a vector.

The Reason (R) is true because the value of magnetic flux can be positive, negative, or zero.

As we know that,

By using Faraday's law, emf induced in coil due to relative motion of coil with field causes increase and decrease of current continuously.

Thus, emf cannot be produced by maintaining large but constant current in a circuit.

According to Faraday's second law,

$\mathrm{emf}=-\frac{d \phi}{d t}$, i.e. rate of change of flux and flux, $\phi=\mathbf{B} \cdot \mathbf{A}=B A \cos \theta$

where, $B$ is magnetic field.

$A$ is area of cross-section.

$\therefore \phi=B A \cos 0 \Upsilon=B A \Rightarrow \phi_{\max }=B A$

$\therefore$ Induced emf is not zero.

Therefore, $A$ is false and $R$ is true.

When plane of coil is perpendicular to the magnetic field, then, $\theta=0\Upsilon$.

Length of solenoid, $l_1=60 \mathrm{~cm}=60 \times 10^{-2} \mathrm{~m}$

Turns per $\mathrm{cm}\left(n_1\right)$ and $\left(n_2\right)$ are 1500 turns/m and 4000 turns$/\mathrm{m}$.

For 1st solenoid, area of cross-section, $ A_1=4 \times 10^{-3} \mathrm{~m}^2$

For 2nd solenoid, area of cross section, $ A_2=2 \times 10^{-3} \mathrm{~m}^2$

$\because$ Mutual inductance is same. (i.e., $M_{12}=M_{21}=M$)

$\begin{aligned} \therefore M & =\left(\mu_0 n_1 n_2 A_2 l\right) \\ & =4 \pi \times 10^{-7} \times 1500 \times 4000 \times 2 \times 10^{-3} \times 60 \times 10^{-2} \\ & =9.04 \times 10^{-3}=9.04 \mathrm{mH} \simeq 9 \mathrm{mH}\end{aligned}$

As we know that, According to Faraday’s law of electromagnetic induction, when we move a conductor in magnetic field region, there will be induced current in the conductor and the same phenomenon is used in case of electric generator where we find electrical energy from mechanical energy due to motion of conductor in magnetic field region.

As, we know that,

$\text { emf induced, } \varepsilon=-\frac{n d \phi}{d t}$

where, $\phi$ is magnetic flux $\phi=B \cdot A$

$n$ is number of turn

$B$ is magnetic field strength

and $A$ is area of cross-section.

$\begin{array}{ll} \therefore & \varepsilon=-\frac{n B A \cos \theta}{t} \\ \Rightarrow & \varepsilon \propto n \end{array}$

$\therefore$ If $n$ increases, emf also increases and this back emf resist motion of magnet. Hence, option (a) is correct.

Given,

Series equivalent inductance,

$L_{\mathrm{eq}_1}=8 \mathrm{H}$

Parallel equivalent inductance,

$L_{\mathrm{eq} 2}=1.5 \mathrm{H}=3 / 2 \mathrm{H}$

Let two inductors used here be $L_1$ and $L_2$, then

$\begin{array}{rlrl} & L_{\mathrm{eq}_1}(\text { series })=L_1+L_2 \\ \therefore & L_{\mathrm{eq}_1}=L_1+L_2 =8\quad .....(\mathrm{i)} \\ \text { and } & \frac{1}{L_{\mathrm{eq}_2}}(\text { parallel })= \frac{1}{L_1}+\frac{1}{L_2} \Rightarrow L_{\mathrm{eq}_2}=\frac{L_1 L_2}{L_2+L_1} \\ \Rightarrow & \frac{3}{2}=\frac{L_1\left(8-L_1\right)}{8} \Rightarrow 12=8 L_1-L_1^2 \\ \Rightarrow & L_1^2-8 L_1+12 =0 \\ \Rightarrow & L_1^2-6 L_1-2 L_1+12 =0 \\ \Rightarrow & L_1\left(L_1-6\right)-2\left(L_1-6\right) =0 \\ \therefore & L_1 =2 \mathrm{H} \text { or } 6 \mathrm{H} \\ \text { and } & L_2 =(8-2) \text { or }(8-6) \\ & =6 \mathrm{H} \text { or } 2 \mathrm{H} \end{array}$

$\therefore$ Difference between inductances is

$\begin{aligned} \left(L_1-L_2\right) \text { or }\left(L_2-L_1\right) & =6-2 \\ & =4 \mathrm{H} \end{aligned}$

According to modified Ampere's law or Maxwell-Ampere's law,

$\nabla \times B=\propto_0\left(J+\varepsilon_0 d E / d t\right)$

where, $B$ is magnetic field,

$J$ is current density

and $\varepsilon_0$ is free-space permittivity.