Electromagnetic Induction

Here, we will use the magnetic flux formula and Faraday's law. Faraday's law relates the induced EMF to the rate of change of magnetic flux, which is appropriate for a loop in a changing magnetic field and changing orientation.

magnetic flux formula: $\phi=\vec{B} \cdot \vec{A}$

$ \Rightarrow \phi=B A \cos \theta $

and the induced EMF is given by Faraday's law: $\quad v=\frac{-d \phi}{d t}$

where, $\phi=$ magnetic flux through the loop

$B=$ magnetic field

$A=$ Area of the loop

$\theta=$ Angle between a perpendicular vector to the area and the magnetic field.

Given: $\quad \vec{B}(t)=B_0(\cos \omega t) \hat{k}$

$ \theta=\omega t $

we know, $\quad \phi=\vec{B} \cdot \vec{A}$

$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& =B A \cos (90-\theta) \\ \phi & =B A \sin \theta\end{aligned}$

$ \begin{aligned} &\begin{aligned} \Rightarrow \phi & =B_0 \cos (\omega t) A \sin (\omega t) \\ \Rightarrow \phi & =\frac{B_0 A}{2} \sin (2 \omega t) \\ \varepsilon=v & =\frac{-d \phi}{d t}=\frac{-d}{d t}\left[\frac{B_0 A}{2} \sin (2 \omega t)\right] \\ & =\frac{-B_0 A}{2}(2 \omega) \cos (2 \omega t) \\ \Rightarrow \varepsilon & =V=-B_0 \omega A \cos (2 \omega t) \end{aligned}\\ &\text { (only for half rotation) } \end{aligned} $

cos function with time period $=\frac{2 \pi}{2 w}$

$ =\frac{\pi}{w} $

Hence, option (A) is correct.

0 to L

$\varepsilon=\mathrm{B} \ell_{\mathrm{ent}} \mathrm{v}=\mathrm{B} \times \frac{\mathrm{x}}{\sqrt{3}} \mathrm{v}$

L to 2 L

$\begin{aligned} & |\operatorname{emf}|=B\left(\frac{L}{\sqrt{3}}-\frac{x_0}{\sqrt{3}}\right) v-B \frac{2 x_0}{\sqrt{3}} v \\ & =\frac{B v L}{\sqrt{3}}-\sqrt{3} B v x_0 \\ & =B v\left[\frac{L}{\sqrt{3}}-\sqrt{3}(x-L)\right] \\ & =\frac{B v}{\sqrt{3}}[L-3 x+3 L] \\ & =\frac{B v}{\sqrt{3}}[4 L-3 x] \\ & \text { at } x=\frac{4 L}{3} \\ & \text { emf }=0 \end{aligned}$

From force equation

$ \begin{aligned} & \mathrm{mg}-\mathrm{Bi} \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\\\ & \mathrm{mg}-\frac{\mathrm{BBi} \ell}{\mathrm{R}} \times \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\\\ & \frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}=\frac{\mathrm{mR}}{\mathrm{B}^2 \ell^2} \frac{\mathrm{dv}}{\mathrm{dt}} \\\\ & \frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}} \int\limits_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}=\int\limits_0^{\mathrm{v}} \frac{\mathrm{dv}}{\left(\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}\right)} \end{aligned} $

Now $\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}}=2$

And $\frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}}=\frac{16 \times \frac{1}{16}}{20 \times 10^{-3} \times 10}=\frac{1}{0.2}=5$

$\begin{aligned} & \therefore 5 \mathrm{t}=[-\ln (2-\mathrm{v})]_0^{\mathrm{v}} \\\\ & -5 \mathrm{t}=\ell \mathrm{n}\left[\frac{2-\mathrm{v}}{\mathrm{v}}\right] \\\\ & \therefore \mathrm{v}=2\left(1-\mathrm{e}^{-5 \mathrm{t}}\right)\end{aligned}$

At $\mathrm{t}=0.2 \mathrm{sec}$

$\begin{aligned} & \mathrm{v}=2\left(1-\mathrm{e}^{-5 \times 0.2}\right) \\\\ & \mathrm{v}=2(1-0.4) \\\\ & \mathrm{v}=1.2 \mathrm{~m} / \mathrm{s}\end{aligned}$

(P) Now at $\mathrm{t}=0.2 \mathrm{sec}$

The magnitude of the induced emf $=\mathrm{E}=\mathrm{Bv} \ell$

$ =4 \times 1.2 \times \frac{1}{4}=1.2 \mathrm{Volt} $

(Q) At $\mathrm{t}=0.2 \mathrm{sec}$, the magnitude of magnetic force $=\mathrm{BI} \ell \sin \theta$

$ \begin{aligned} & =\mathrm{B} \times \frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}} \times \ell \times \sin 90^{\circ} \\\\ & =\frac{4 \times 4 \times \frac{1}{4} \times 1.3 \times \frac{1}{4}}{10}=0.12 \text { Newton } \end{aligned} $

(R) At t $=0.2 \mathrm{sec}$, the power dissipated as heat

$ \begin{aligned} & P=i^2 R=\frac{v^2}{R}=\frac{1.2 \times 1.2}{10} \\\\ & P=0.144 \text { watt } \end{aligned} $

(S) Magnitude of terminal velocity

At terminal velocity, the net force become zero

$ \begin{aligned} & \therefore \mathrm{mg}=\mathrm{Bi} \ell \\\\ & \mathrm{mg}=\mathrm{B} \times \frac{\mathrm{B} \ell \mathrm{v}_{\mathrm{t}}}{\mathrm{R}} \times \ell \\\\ & \therefore \mathrm{v}_{\mathrm{T}}=\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}} \\\\ & \mathrm{v}_{\mathrm{T}}=2 \mathrm{~m} / \mathrm{s} \end{aligned} $

Hence, Answer is (D)

Since $\phi_B$ increases in the downward direction, according to Lenz's law, the induced current flows in such a manner to create an upward going flux, that is, a flux in anticlockwise direction. Since the outer loop has a bigger enclosed area than inner loop, the induced current flow is as follows:

$\bullet$ Induced current, $I_2$: Flows from d to c

$\bullet$ Induced current, $I_1$: Flows from a to b

In maglev system, when train is in levitation, there is no friction between the train and the track. Hence, no power consumption in this case and overall power consumption reduces.

Disadvantages of Maglev train is that as it slows down, electromagnetic forces decreases and it becomes difficult to keep levitated and as it moves forward according to Lenz's law there is electromagnetic drag force.

As the train moves, due to induced magnetic field train elevates.

According to the question, radius of the solenoid is $r=a$,

Number of turns per unit length $n$,

Radius of the cylinder shell is $\mathrm{R}$,

Thickness of shell is $d$, length is L,

Resistivity of cylindrical shell is $\rho$.

The expression of the current through solenoid is given as,

$\mathrm{I}=\mathrm{I}_{0} \sin \omega t$

In the given case the magnetic field outside the solenoid is zero. Therefore, magnetic field inside the solenoid is $\mu_{0} n_{i}$

Now, magnetic flux can be written as $\phi=\mu_{0} n \mathrm{I} \times \pi a^{2}$ Therefore, induced emf can be written as

$\begin{aligned} e & =-\frac{d \phi}{d t} \\ e & =-\frac{d}{d t}\left(\mu_{0} n \mathrm{I} \pi a^{2}\right) \end{aligned}$

As current through the solenoid is given as

$\begin{aligned} & \mathrm{I}=\mathrm{I}_{0} \sin w t \\ & \Rightarrow \quad e=-\frac{d}{d t}\left(\mu_{0} n \mathrm{I}_{0} \sin \omega t \pi a^{2}\right) \\ & =-\left(\mu_{0} n \pi a^{2} \mathrm{I}_{0}\right) \frac{d}{d t} \sin \omega t \\ & \Rightarrow \quad e=\left(\mu_{0} n \pi a^{2}\right)\left(\mathrm{I}_{0} l_{0} \omega \cos \omega t\right) \end{aligned}$

Resistance of the cylindrical vessel,

$\mathrm{R}_{c}=\frac{\rho}{\mathrm{S}}=\frac{\rho(2 \pi \mathrm{R})}{\mathrm{L} d}$

$\therefore$ Induced current, $\mathrm{I}=\frac{e}{\mathrm{R}_{c}}$

$=\frac{\left(\mu_{0} n \pi a^{2}\right)\left(\mathrm{I}_{0} \omega \cos \omega t\right)}{\frac{\rho(2 \pi \mathrm{R})}{\mathrm{L} d}}$

$\mathrm{I}=\frac{\mu_{0} \mathrm{~L} d n a^{2} i_{0} \omega \cos \omega t}{2 \rho R}$

Here, as the loop enters the field, the magnetic flux through the loop is changing. Changing magnetic flux, according to Faraday's law, induces an EMF in the loop. As the loop has some resistance
So the loop gets an induced current and

The magnetic force arises from the induced current interacting with the field.

Flux : $\phi=B_0$. (Area in Field)

$ \Rightarrow \phi=B_0 \cdot L y $

EMF: $\quad V=\frac{-d \phi}{d t}=-\frac{d}{d t}\left(B_0 L y\right)$.

$ \begin{aligned} & \Rightarrow \quad v=-B_0 L \frac{d y}{d t} \\ & \Rightarrow \quad v=-B_0 L v \end{aligned} $

current $(v>0$, so current direction based on Lenz's (law): $I=\frac{V}{R}=\frac{-B_0 v L}{R}$

Force: current in the bottom edge (in field) interacts with $\vec{B}$

so force on a segment: $F=B_0 I L$

$ \begin{aligned} & \Rightarrow F=B_0\left(\frac{-B_0 v L}{R}\right) L \\ & \Rightarrow F=\frac{-B_0^2 L^2 v}{R} \end{aligned} $

$ \text { Given, } K=\frac{B_0^2 L^2}{R M} $

$ \begin{aligned} \text { or } \varepsilon & =-B_0 v L \\ I & =\frac{-B_0 v L}{R} \\ F & =B_0 I L=B_0\left(\frac{-B_0 v L}{R}\right) L \end{aligned} $

$ \Rightarrow \frac{m d v}{d t}=\frac{-\left (B_0^2 L^2\right) v}{R} \quad\left[\begin{array}{ll} \text { as } & F=m a \\ & =m \frac{d v}{d t} \end{array}\right] $

$ \begin{aligned} & \Rightarrow \text { } \frac{mvd v}{d x}=\frac{-\left(B_0^2 L^2\right)}{R} {v} \\ & \Rightarrow \frac{d v}{d x}=-\frac{B_0^2 L^2}{M R} \\ & \Rightarrow \int_{v_0}^v d v=-K \int_0^{x} dx\\ & \Rightarrow v-v_0=-K x \\ & \text { If } v_0=1.5 \mathrm{KL} \text { then for } x=L \end{aligned} $

$ \begin{aligned} & \Rightarrow v-1.5 \mathrm{KL}=-K L \\ & \Rightarrow \quad v=1.5 \mathrm{KL}-\mathrm{KL}=0.5 \mathrm{KL} \\ & \Rightarrow \quad v=0.5 \mathrm{KL} \end{aligned} $

So, the loop will have some velocity, and it will not stop.

If the loop completely gets inside then $\Phi$ becomes constant so $\frac{d \Phi}{d t}=\varepsilon=0, \quad I=0$
i.e., no current flows and no force acts.

if $v_0=\frac{K L}{10}$ then for $v=0$

$ \begin{aligned} & 0+\frac{K L}{10}=-k x \Rightarrow x=\frac{L}{10} \\ & \frac{d v}{d t}=\frac{-B_0^2 L^2}{M R} v \\ & \Rightarrow \int_{v_0}^v \frac{d v}{v}=-k \int_0^t d t \\ & \Rightarrow[\ln |v|]_{v_0}^v=-k t \\ & \Rightarrow \ln \left|\frac{v}{v_0}\right|=-k t \\ & \Rightarrow t=\frac{1}{K} \ln \left|\frac{v_0}{v}\right| \\ & \text { for } v=0, t=\frac{1}{k} \ln (\infty)=\infty \end{aligned} $

So it will never stop.

If $v_0=3 \mathrm{KL}$ then

$v$ at $x=L$,

$v-3 K L=-K L$

$ \Rightarrow v=3 K L-K L=2 K L $

So time, $t=\frac{1}{k} \ln \left|\frac{v_0}{v}\right|$

$ \begin{aligned} & \quad t=\frac{1}{k} \ln \left|\frac{3 K L}{2 K L}\right| \\ & \Rightarrow t=\frac{1}{k} \ln \left(\frac{3}{2}\right) \end{aligned} $

Hence, Options (B) and (D) are correct.

Motional emf across the length dy is,

$d\varepsilon = B{V_0}\,dy = {B_0}\left[ {1 + {{\left( {{y \over L}} \right)}^\beta }} \right]{V_0}dy$

$ \therefore $ $\varepsilon = \int\limits_0^L {{B_0}} \left( {1 + {{\left( {{y \over L}} \right)}^\beta }} \right){V_0}dy$

$ = {B_0}{V_0}\left[ {1 + {1 \over {\beta + 1}}} \right]$

emf in loop is proportional to L for given value of $\beta $,

$\beta = 0,\,\varepsilon = 2{B_0}{V_0}L$

$\beta = 2,\,\varepsilon = {B_0}{V_0}L\left[ {1 + {1 \over 3}} \right] = {4 \over 3}{B_0}{V_0}L$

The length of projection of the wire Y = X of length $\sqrt 2 L$ on the y-axis is thus, the answer remain unchanged.

Therefore, correct options are (a), (b) and (d)

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

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

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

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

Let i₁ be the current through L₁, i₂ be the current through L₂ and i be the current through R.

The inductors $ L_1 $ and $ L_2 $ are connected in parallel. Therefore, the self-induced emf in $ L_1 $ and $ L_2 $ are equal, which can be expressed as:

$ \varepsilon_1 = -L_1 \frac{d i_1}{dt} = \varepsilon_2 = -L_2 \frac{d i_2}{dt} $

Upon integrating this equation, we get :

$ L_1 i_1 = L_2 i_2 $ ............(1)

This indicates that the ratio of the currents through $ L_1 $ and $ L_2 $ is constant.

Using Kirchhoff's current law at junction A, we have :

$ i = i_1 + i_2 $ .............(2)

At the moment the switch is closed (t = 0), the inductors act as open circuits (since the inductive reactance $ X_L = \omega L $ approaches infinity as $ \omega \to \infty $). Therefore, initially :

$ i = i_1 + i_2 = 0 $

and the current through the resistor is :

$ i = i_1 + i_2 = 0 $

After a long time, the inductive reactance of the inductors becomes zero. Applying Kirchhoff's loop law, the current through the resistor $ R $ is :

$ i = \frac{V}{R} $

Substituting $ i = \frac{V}{R} $ into equation (2) and solving equations (1) and (2), we find :

$ i_1 = \frac{V}{R} \frac{L_2}{L_1 + L_2} $

$ i_2 = \frac{V}{R} \frac{L_1}{L_1 + L_2} $

Consider the coordinate axes shown in the figure. The magnetic field is into the paper, i.e., $\overrightarrow B = B\widehat k$. Let loop 1 be the loop of area A and loop 2 be the loop of area 2A. At t = 0, both loops are in the plane of the paper. They start rotating about the x-axis with a constant angular velocity $\overrightarrow \omega = \omega \widehat i$. Let the area vectors of loop 1 and 2 be ${\overrightarrow A _1}$ and ${\overrightarrow A _2}$, respectively. At t = 0, ${\overrightarrow A _1}$ = $A\widehat k$ and ${\overrightarrow A _2} = 2A\widehat k$ (we chose area into the plane of the paper as positive). In time t, the loops rotate by an angle $\theta$ = $\omega$t. Their area vectors make an angle $\theta$ = $\omega$t with z-axis (direction of the magnetic field). Thus, the magnetic fluxes passing through the loop 1 and the loop 2 at time t are given by

$\phi$₁ = $\overrightarrow B .{\overrightarrow A _1}$ = B A₁ cos$\theta$ = B A cos $\omega$t,

$\phi$₂ = $\overrightarrow B .{\overrightarrow A _2}$ = B A₂ cos$\theta$ = 2B A cos $\omega$t.

By Faraday's law of electromagnetic induction, induced emfs in the loop 1 and in the loop 2 are given by

${\varepsilon _1} = - {{d{\phi _1}} \over {dt}} = B\,A\omega \sin \omega t$,

${\varepsilon _2} = - {{d{\phi _2}} \over {dt}} = 2B\,A\omega \sin \omega t$.

The rate of change of flux is maximum when $\omega$t = $\pi$/2 i.e., when the plane of the loops is perpendicular to the plane to the paper.

Let T = 2$\pi$/$\omega$ be the time period to complete one revolution. The fluxes $\phi$₁ and $\phi$₂ decrease with time if t $\in$ (0, T/4). By Lenz's law, the induced current should oppose reductions in $\phi$₁ and $\phi$₂. Thus, induced current in both the loops should be clockwise. Hence, the polarity of 'equivalent batteries' are as shown in the figure. These batteries are connected with reverse polarity. Thus, net emf induced due to both the loop is

$\varepsilon = {\varepsilon _2} - {\varepsilon _1} = B\,A\omega \sin \omega t$.

Note that the net emf $\varepsilon $ is proportional to the difference in areas of two loops and $\left| \varepsilon \right| = \left| {{\varepsilon _1}} \right|$.

We discuss the following cases :

$\bullet$ For x = 0 $\to$ L : The induced emf in the loop is

$\varepsilon = v{B_0}L$

Therefore, the induced current is

$I = {{v{B_0}L} \over R}$ ........ (1)

As the flux is increasing, the current passes in counterclockwise direction and oppose the existing magnetic field (i.e., Lenz's law).

Let us consider the loop which is labelled as shown in the following figure:

Force on the side CD of the loop is

$\overrightarrow F = i(\overrightarrow L \times \overrightarrow B ) = {{v{B_0}L} \over R}[L\widehat j \times {B_0}( - \widehat k)]$

$\overrightarrow F = - {{vB_0^2{L^2}} \over R}\widehat i$ (towards left) ........ (2)

$ \Rightarrow {{mdv} \over {dt}} = - {{B_0^2{L^2}} \over R}v$

$ \Rightarrow {{mdv} \over {dx}}.{{dx} \over {dt}} = - {{ - B_0^2{L^2}} \over R}v$

$ \Rightarrow \int\limits_{{v_0}}^v {dv = - {{B_0^2{L^2}} \over {mR}}\int\limits_0^x {dx} } $ (Since ${{dx} \over {dt}} = v$)

$ \Rightarrow v = {v_0} - {{B_0^2{L^2}} \over {mR}}x$ ....... (3)

From Eqs. (1) and (3), we get

$I = {{{v_0}{B_0}L} \over R} - {{B_0^3{L^3}} \over {m{R^2}}}x$ ....... (4)

From Eqs. (2) and (3), we get

$\overrightarrow F = - {{{v_0}B_0^2{L^2}} \over R} + {{B_0^4{L^4}} \over {m{R^2}}}x$

$\bullet$ For x = L $\to$ 3L: Here, there is no change in flux and hence there is no induced current and thereby there is no force and no change in velocity.

$\bullet$ For x = 3L $\to$ 4L :

As the flux is decreasing, the current passes in clockwise direction and supports the existing magnetic field (i.e., Lenz's law).

Hence, option (A) is incorrect.

Consequently, the magnetic force experienced by the arm AB of the loop is

$\overrightarrow F = i(\overrightarrow L \times \overrightarrow B ) = {{v{B_0}L} \over R}[L\widehat j \times {B_0}( - \widehat k)]$

$ = - {{vB_0^2{L^2}} \over R}\widehat i$ (towards left)

Hence, option (B) is incorrect.

Similarly, now we have

$\int\limits_{{v_{3L}}}^v {dv = - {{B_0^2{L^2}} \over {mR}}\int\limits_{3L}^x {dx \Rightarrow v = {v_{3L}} - {{B_0^2{L^2}} \over {mR}}\left( {x - 3L} \right)} } $

When the loop is completely within the magnetic field, the force and current is zero and velocity is constant. However, while coming out, the current passes in clockwise direction, the force exists in backwards and the velocity decreases linearly.

The flux passing through the triangular wire if ' $i$ ' current flows through the infinitely long conducting wire.


$ \begin{aligned} \frac{d \mathrm{I}}{d t} & =10 \mathrm{As}^{-1} \\\\ \phi & =\int_0^{0.1} \frac{\mu_0 i}{2 \pi x} \times 2 x d x \\\\ \phi & =\frac{\mu_0 i}{10 \pi}=\mathrm{M} i \end{aligned} $

$ \therefore \mathrm{M}=\frac{\mu_0}{10 \pi} $

$ \text { Induced emf in the wire }=\frac{\mathrm{M} d i}{d t}=\frac{\mu_0}{10 \pi} \times 10=\frac{\mu_0}{\pi} $

According to Lenz's law, the induced current opposes the flux change that creates it. Since the flux is increasing, the induced emf opposes the existing magnetic field and the current flows from left to right in the straight wire, that is, along the hypotenuse current. That is, there is a repulsive force between the wire and the loop.


Also, if the loop is rotated about the wire, there is no change in the flux through the wire, so no extra induced emf is generated in the wire.

The magnetic field due to an infinitely long conductor is circumferential.

Since this conductor is placed along the diameter of circular loop (see figure), the total magnetic flux through the circular wire loop is always zero.

Induced emf is same in both the rings:

$I = {e \over R} = {{\rho A} \over {\rho l}}$

$I \propto {1 \over \rho } \Rightarrow q \propto {1 \over \rho }$ .... (1)

Impulse is

$J = \int {Bil\,dt = mv = Bl\int {I\,dl = mv} } $

That is, $J = Blq = mv \Rightarrow v\left( {{q \over m}} \right)$ ..... (2)

and ${v^2} \propto h$ ...... (3)

From Eqs. (1), (2) and (3), we get

$mv \propto {1 \over \rho }$

$m\sqrt h \propto {1 \over \rho }$

$m\rho \propto {1 \over {\sqrt h }}$

Since ${h_A} > {h_B}$ and for ${m_A} = {m_B},\rho \sqrt h = $ Constant. Therefore, ${\rho _A} < {\rho _B}$. Also if ${m_A} < {m_B}$ and ${\rho _A} < {\rho _B}$,

${m_A}{\rho _A} < {m_B}{\rho _B}$

$ \Rightarrow {h_A} > {h_B}$ (already given)