Electromagnetic Induction

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.

According to given condition, Magnetic energy stored in solenoid $X=$ Magnetic energy stored in solenoid $Y$

$ \begin{array}{ll} \Rightarrow & U_X=U_Y \\ \Rightarrow & \frac{B_X^2 V_X}{2 \mu_0}=\frac{B_Y^2 V_Y}{2 \mu_0} \\ & {\left[\because U=\frac{B^2 V}{2 \mu_0}, V=\text { volume }\right]} \\ & \\ & B_X^2 V_X=B_Y^2 V_Y \end{array} $

$ \begin{aligned} & \Rightarrow \quad B_X^2 A_X L_X=B_Y^2 A_Y L_Y \\ & \Rightarrow \quad \frac{B_X^2}{B_Y^2}=\frac{A_Y}{A_X} \times \frac{L_Y}{L_X}=\frac{2 A_X}{A_X} \times \frac{2 L_X}{L_X} \\ & \Rightarrow \quad \frac{B_X^2}{B_Y^2}=4 \Rightarrow \frac{B_X}{B_Y}=\sqrt{4} \\ & \Rightarrow \quad \frac{B_X}{B_Y}=2 \Rightarrow B_X: B_Y=2: 1 \end{aligned} $

Self-induced emf in coil is given by

$ E=L(d I / d t) $

Here, $E=200 \mathrm{~V}, \frac{d I}{d t}=\left(\frac{10-0}{1.5}\right) \mathrm{A} / \mathrm{s}$

So, self-inductance of coil,

$ \begin{aligned} L & =\frac{E}{d I / d t}=\frac{200}{\left(\frac{10-0}{1.5}\right)} \\ \Rightarrow L & =\frac{200 \times 1.5}{10}=30 \mathrm{H} \end{aligned} $

We can apply Faraday's law of electromagnetic induction to solve this problem. Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.

The cube is moving through a magnetic field, so it's behaving like a conductor moving through a magnetic field. The induced emf or voltage can be calculated by using the formula:

emf = B $ \times $ v $ \times $ d

Where:

B is the magnetic field strength,

v is the velocity of the conductor, and

d is the length of the conductor perpendicular to the direction of motion and magnetic field.

In this case, the cube is moving in the y-direction and the magnetic field is in the z-direction. So, the faces perpendicular to the x-axis are involved. The length of the conductor (d) perpendicular to the motion and magnetic field is the edge length of the cube, which is 2 cm or 0.02 m.

So, plugging the given values into the formula:

emf = 0.1 T $ \times $ 6 m/s $ \times $ 0.02 m = 0.012 V = 12 mV

So, the potential difference between the two faces of the cube perpendicular to the x-axis is 12 mV.

Therefore, Option B is correct.

At terminal velocity, net force on rod = $mg\sin \theta $

$ \Rightarrow mg\sin \theta = iBl$

Now, $IBl = \left( {{{Bvl} \over R}} \right)Bl = {{{B^2}{l^2}v} \over R}$

$ \Rightarrow mg\sin \theta = {{{B^2}{l^2}v} \over R} \Rightarrow v = {{mgR\sin \theta } \over {{B^2}{l^2}}}$

We know that electric flux $\phi = \overrightarrow B \,.\,\overrightarrow A $

$ \Rightarrow \phi = BA\cos \omega t$

Now, $B = {{{\mu _0}} \over 2}{I \over R}$ is magnetic field due to circular coil of radius R and $A = \pi {r^2}$ is area of circular coil of radius r. Therefore,

$\phi = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t$

Now induced emf $\varepsilon = {{ - d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t} \right)$

$ \Rightarrow \varepsilon = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\sin \omega t$

${I_1} = {V \over R}\left( {1 - {e^{ - {{tR} \over L}}}} \right)$

${I_2} = {V \over R}\left( {1 - {e^{ - {{tR} \over {2L}}}}} \right)$

From principle of superposition,

$I = {I_1} - {I_2} \Rightarrow I = {V \over R}{e^{ - {{tR} \over {2L}}}}\left( {1 - {e^{ - {{tR} \over {2L}}}}} \right)$ ...... (i)

I is maximum when ${{dI} \over {dt}} = 0$, which gives

${e^{ - {{tR} \over {2L}}}} = {1 \over 2}$ or $t = {{2L} \over R}\ln 2$

Substituting this time in Eq. (i), we get

${I_{\max }} = {V \over {4R}}$