Electromagnetic Induction

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)

Given,

Velocity of conducting rod $=v$

Magnetic field $=B$

Induced current $=I$

For induced current direction, we use Fleming's right hand rule.

It states that if thumb, forefinger and middle finger of right hand are mutually perpendicular to each other, then if fore finger indicate the direction of external magnetic field ( $B$ ) and thumb shows direction of motion of conductor, then middle finger will show the direction of induced current and hence direction of emf.

Hence, from diagram and applying the Fleming's right hand rule, the direction of magnetic field is perpendicular to plane of paper and into the paper.

Metal detector works on the principle of electromagnetic induction. A metal detector produces a time varying magnetic field from the coil inside. When the hidden metal is exposed to varying magnetic field, it experience a change in flux. As a result EMF in induced and eddy current is produced. Since, charge in motion create magnetic field, eddy current produces its own magnetic field. This field is detected by metal detector.

Given,

Copper disc radius, $r=0.1 \mathrm{~m}$

Angular rotation, $n=10 \mathrm{rev} / \mathrm{s}$

Magnetic field, $B=0.1 \mathrm{~T}$

Let a small radial portion of $d x$ at a distance $X$ from centre of disc.

$ \begin{aligned} &\text { Velocity of this element is }\\ &v=\omega X \end{aligned} $

Emf induced across this element.

$ d e=v B d x $

or $\omega x B d x$

Total emf induced across radius $r$,

$ \begin{aligned} E & =\frac{1}{2} B \omega r^2, \text { where } \omega=2 \pi n=2 \pi \times 10 \\ & =20 \pi \mathrm{rad} / \mathrm{s} \end{aligned} $

Put the values in above equation,

$ \begin{aligned} E & =\frac{1}{2} \times 0.1 \times 20 \pi \times 0.1 \times 0.1 \\ & =10 \pi \times 10^{-3} \mathrm{~V} \\ E & =10 \pi \mathrm{mV} \end{aligned} $

Self inductance of a coil is given as

$ L=\frac{\mu_0 \pi n^2 r}{2} $

Where, $\mu$ is permeability $n$ is no. of turns and $A$ is area of coil Thus, self inductance of a coil depends on size, shape of the coil, and number of turns in it.

$ \begin{aligned} &\text { Induced emf is given by }\\ &\begin{aligned} & \begin{aligned} e & =\frac{-d \phi}{d t}=-\frac{d}{d t} \cdot(B A \cos \theta) \\ \Rightarrow \quad e & =+B A \sin \theta \Rightarrow \frac{e_1}{e_2}=\frac{\sin \theta_1}{\sin \theta_2} \\ \text { Here, } \theta_1 & =\left(90^{\circ}-60^{\circ}\right)=30^{\circ} \\ \theta_2 & =180^{\circ}-120^{\circ}=60^{\circ} \\ \frac{e_1}{e_2} & =\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 / 2}{\sqrt{3} / 2}=\frac{1}{\sqrt{3}} \end{aligned} \end{aligned} \end{aligned} $

Speed of aeroplane,

$ \begin{aligned} v & =540 \mathrm{~km} / \mathrm{h} \\ & =540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} \\ & =150 \mathrm{~m} / \mathrm{s} \end{aligned} $

Length of wing span, $l=20 \mathrm{~m}$

$ \begin{aligned} B_H & =25 \sqrt{3} \times 10^{-4} \mathrm{~T} \\ \delta & =30^{\circ} \end{aligned} $

Potential difference developed between the ends of the wing

$ \begin{aligned} \varepsilon & =B_V v l=\left(B_H \tan \delta\right) v l \\ & =2.5 \sqrt{3} \times 10^{-4} \times \tan 30 \times 150 \times 20 \\ & =2.5 \sqrt{3} \times 10^{-4} \times \frac{1}{\sqrt{3}} \times 3 \times 10^3 \\ & =0.75 \mathrm{~V} \end{aligned} $

Given the problem, let's break down the information provided and the calculations involved.

Given Data:

Speed of the aeroplane: $360 \, \text{km/h}$

Convert the speed to meters per second:

$ 360 \, \text{km/h} = 100 \, \text{m/s} $

Wing span of the aeroplane: $4 \, \text{m}$

Vertical component of Earth's magnetic field: $0.5 \times 10^{-4} \, \text{T}$

Formula Used:

The induced electromotive force (emf) generated across the wing span of the aeroplane due to its movement in the Earth's magnetic field is calculated using the formula:

$ e = B \cdot L \cdot v $

Where:

$e$ is the induced emf

$B$ is the magnetic field component ($0.5 \times 10^{-4} \, \text{T}$)

$L$ is the wing span ($4 \, \text{m}$)

$v$ is the speed of the aeroplane ($100 \, \text{m/s}$)

Calculation:

Substitute the values into the formula:

$ e = 0.5 \times 10^{-4} \times 4 \times 100 $

Solving this gives:

$ e = 2 \times 10^{-2} \, \text{V} = 20 \times 10^{-3} \, \text{V} $

Result:

Thus, the induced emf across the ends of the wings is $20 \times 10^{-3} \, \text{V}$.

Explanation:

To determine the induced electromotive force (emf) in the rim of the bicycle wheel, we need to consider the following information:

Radius of the Rim, $ r $: The radius is given as $ 40 \, \text{cm} $, which converts to $ 0.4 \, \text{m} $.

Angular Speed, $ \omega $: The wheel is rotating with an angular speed of $ 20 \, \text{rad/s} $.

Horizontal Component of Earth's Magnetic Field: Given as $ 0.26 \, \text{G} $, which is $ 0.26 \times 10^{-4} \, \text{T} $.

The formula for the induced emf is given by:

$ e = l(\mathbf{v} \times \mathbf{B}) $

Where:

$ e $ is the induced emf.

$ l $ is the length of the conductor or loop.

$ \mathbf{v} $ is the velocity of the loop.

$ \mathbf{B} $ is the magnetic field.

Since the plane of the wheel's revolution is parallel to the horizontal component of Earth's magnetic field, the cross product $ \mathbf{v} \times \mathbf{B} $ becomes zero. This is because the velocity vector of the wheel's rim is parallel to the direction of the magnetic field.

Thus, the induced emf is zero:

$ e = 0 $

Therefore, no emf is induced in this specific orientation of the wheel because the necessary conditions for inducing emf (a changing magnetic field perpendicular to the motion) are not met.

When the current $i$ through a solenoid increases at a constant rate, the analysis of the induced current involves understanding the behavior of magnetic field, magnetic flux, and their rates of change:

Magnetic Field through the Solenoid:

The expression for the magnetic field $ B $ within a solenoid is given by:

$ B = \mu_0 n I $

where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current.

Rate of Change of Magnetic Field:

The rate of change of the magnetic field with respect to time is:

$ \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} $

Since $\frac{dI}{dt}$ is constant (the current is increasing at a constant rate), $\frac{dB}{dt}$ is also constant.

Magnetic Flux through the Solenoid:

Magnetic flux $\phi$ through the solenoid is given by:

$ \phi = B \cdot A $

where $A$ is the cross-sectional area of the solenoid.

Rate of Change of Magnetic Flux:

The rate of change of magnetic flux is expressed as:

$ \frac{d\phi}{dt} = A \frac{dB}{dt} $

Given that $\frac{dB}{dt}$ is constant, $\frac{d\phi}{dt}$ is also constant.

Induced Current:

According to Lenz’s law, the induced current will act to oppose the change in magnetic flux. Therefore, even though the induced current is constant, it arises due to the constant rate of change of flux, working in opposition to the increase in the original current $i$.

Thus, the induced current is constant and occurs in a direction opposite to the original increasing current $i$.

To determine the mutual inductance of a pair of adjacent coils, we start with the given information:

The change in current ($\Delta I$) is from 10 A to 2 A, so $\Delta I = 10 \, \text{A} - 2 \, \text{A} = 8 \, \text{A}$.

The time interval ($\Delta t$) for this change is 0.2 s.

The induced electromotive force (emf) is 120 V.

We use Faraday's law of mutual induction to find the mutual inductance $M$. The formula for the induced emf ($E$) can be derived from Faraday's law as:

$ E = \frac{M \cdot \Delta I}{\Delta t} $

Substitute the known values into the equation:

$ 120 = \frac{M \times 8}{0.2} $

Solving for $M$:

$ 120 = \frac{M \times 8}{0.2} $

$ 120 = M \times 40 $

$ M = \frac{120}{40} = 3 \, \text{H} $

Thus, the mutual inductance of the pair of coils is 3 H.

$B = {{n{\mu _0}I} \over {2R}}$

${B_1} = {{2{\mu _0}I} \over {2{R_1}}}$

${B_2} = {{5{\mu _0}I} \over {2{R_2}}}$

${R_2} = {{2{R_1}} \over 5}$

$ \Rightarrow {{{B_2}} \over {{B_1}}} = {5 \over 2} \times {{{R_1}} \over {{R_2}}} = {{25} \over 4}$

We know $\phi = Mi$

Let i current be flowing in the larger loop

$ \Rightarrow \phi \simeq \left[ {4 \times {{{\mu _0}i} \over {4\pi (L/2)}}[\sin 45^\circ + \sin 45^\circ ]} \right] \times $ Area

$ = {{2\sqrt 2 {\mu _0}i} \over {\pi L}} \times {I^2}$

$ \Rightarrow M = {\phi \over i} = {{2\sqrt 2 {\mu _0}{I^2}} \over {\pi L}}$

The electrical power dissipated due to the current induced in a coil placed in a time-varying magnetic field can be determined by considering Faraday's Law of Induction and the resistance of the coil.

Faraday's Law of Induction

The induced EMF ($\mathcal{E}$) in a coil with $N$ turns experiencing a time-varying magnetic flux ($\Phi_B$) is given by:

$ \mathcal{E} = -N \frac{d\Phi_B}{dt} $

Resistance of the Coil

The resistance ($R$) of a wire depends on its length ($L$) and cross-sectional area ($A$) as well as the resistivity ($\rho$) of the material:

$ R = \frac{\rho L}{A} $

For a coil of radius $r$ and with wire of radius $a$, assuming the wire is wound tightly with a length approximated by the circumference of the coil multiplied by the number of turns, we have:

$ L \approx 2\pi r N $

And the cross-sectional area of the wire is given by:

$ A = \pi a^2 $

Thus, the resistance becomes:

$ R \approx \frac{\rho \cdot 2\pi r N}{\pi a^2} = \frac{2\rho r N}{a^2} $

Power Dissipated

The electrical power ($P$) dissipated in the coil is related to the induced current ($I$) and the resistance ($R$):

$ P = I^2 R $

Using Ohm's Law, the induced current $I$ can be expressed as:

$ I = \frac{\mathcal{E}}{R} $

Substituting the expressions for $\mathcal{E}$ and $R$:

$ I = \frac{N \frac{d\Phi_B}{dt}}{\frac{2\rho r N}{a^2}} = \frac{a^2}{2\rho r} \cdot \frac{d\Phi_B}{dt} $

Hence, the power dissipated:

$ P = \left(\frac{a^2}{2\rho r} \cdot \frac{d\Phi_B}{dt}\right)^2 \cdot \frac{2\rho r N}{a^2} $

$ P = \frac{a^4}{4\rho^2 r^2} \left( \frac{d\Phi_B}{dt} \right)^2 \cdot \frac{2\rho r N}{a^2} $

$ P = \frac{a^2}{2\rho r} \cdot N \left( \frac{d\Phi_B}{dt} \right)^2 $

Case when the number of turns is halved and the radius of the wire is doubled

Halving the number of turns:

$ N' = \frac{N}{2} $

Doubling the radius of the wire:

$ a' = 2a $

Substituting these into the power formula:

$ P' = \frac{(2a)^2}{2\rho r} \cdot \frac{N}{2} \left( \frac{d\Phi_B}{dt} \right)^2 $

$ P' = \frac{4a^2}{2\rho r} \cdot \frac{N}{2} \left( \frac{d\Phi_B}{dt} \right)^2 $

$ P' = \frac{4a^2}{2\rho r} \cdot \frac{N}{2} \left( \frac{d\Phi_B}{dt} \right)^2 $

$ P' = 2 \left( \frac{a^2}{2\rho r} \cdot N \left( \frac{d\Phi_B}{dt} \right)^2 \right) $

$ P' = 2P $

Thus, the electrical power dissipated in the coil would be:

Option D

Doubled

Self inductances are in series but their mutual inductances are linked oppositely so equivalent self inductance

L = L₁ + L₂ $-$ M $-$ M = L₁ + L₂ $-$ 2M

$Emf = {1 \over 2}B\omega {l^2}$

$ = {1 \over 2} \times 0.2 \times {10^{ - 4}} \times 5 \times {1^2}$ V

= 0.5 $\times$ 10^$-$4 V

= 50 $\mu$V

$\phi=5 \mathrm{t}^3+4 \mathrm{t}^2+2 \mathrm{t}-5$

$|\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=15 \mathrm{t}^2+8 \mathrm{t}+2$

At $\mathrm{t}=2,|\mathrm{e}|=15 \times 2^2+8 \times 2+2$

$\Rightarrow \mathrm{e}=78 \mathrm{~V} $

$\Rightarrow \mathrm{I}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{78}{5}=15.60$

Given, length of each spokes of wheel, $l=40 \mathrm{~cm}=0.4 \mathrm{~m}$

Angular velocity, $\omega=2 \pi f$

$ \begin{aligned} & =2 \pi \times \frac{180}{60} \mathrm{rad} / \mathrm{s} \\ & =6 \pi \mathrm{rad} / \mathrm{s} \end{aligned} $

Horizontal component of earth's magnetic field,

$ H_{\mathrm{e}}=0.4 \mathrm{G}=0.4 \times 10^{-4} \mathrm{~T} $

$ \begin{aligned} & \therefore \text { Induced emf, } e=\frac{1}{2} B \omega L^2 \\ & \qquad \begin{aligned} = & \frac{1}{2} \times 0.4 \times 10^{-4} \times 6 \pi \times(0.4)^2 \\ = & 1.2 \pi \times 10^{-4} \times 0.16 \\ = & 0.192 \pi \times 10^{-4} \mathrm{~V} \\ = & 192 \pi \times 10^{-7} \mathrm{~V} \end{aligned} \end{aligned} $

Radius of metal disc,

$ R=30 \mathrm{~cm}=0.3 \mathrm{~m} $

Angular velocity, $\omega=100 \mathrm{rad} / \mathrm{s}$

Magnetic field, $B=4 \mathrm{mT}$

$ =4 \times 10^{-3} \mathrm{~T} $

Magnitude of potential difference between the centre and the rim of the disc.

$ \begin{aligned} e & =\frac{1}{2} B \omega R^2 \\ & =\frac{1}{2} \times 4 \times 10^{-3} \times 100 \times(0.3)^2 \\ & =2 \times 10^{-3} \times 100 \times 0.09 \\ & =18 \times 10^{-3} \mathrm{~V}=18 \mathrm{mV} \end{aligned} $

The given situation is shown below.

Magnetic field,

$ B=2 \mathrm{~T} $

Area of the triangular loop,

$ \begin{aligned} A & =\frac{1}{2} \times P Q \times P R \\ & =\frac{1}{2} \times 2 \times 1=1 \mathrm{~cm}^2 \\ & =10^{-4} \mathrm{~m}^2 \end{aligned} $

∴ Magnetic flux through triangular loop,

$ \phi=B A=2 \times 10^{-4} \mathrm{~Wb} $

Area of wire loop, $A=0.2 \mathrm{~m}^2$

Resistance, $R=20 \Omega$

Magnetic field, $B=0.25 \mathrm{~T}$

According to Faraday's law of electromagnetic induction.

$ \begin{aligned} & \text { Induced emf, } e=\frac{-d \phi}{d t} \\ & \qquad \begin{aligned} & =\frac{-d}{d t} B A=-A \frac{d B}{d t} \\ & =-A \frac{\Delta B}{\Delta t}=-A \frac{\left(B_2-B_1\right)}{10^{-4}} \\ & =-0.2 \frac{(0-0.25)}{10^{-4}}=\frac{0.05}{10^{-4}} \\ \Rightarrow \quad e & =500 \mathrm{~V} \end{aligned} \\ & \begin{aligned} \end{aligned} \\ & \end{aligned} $

and induced current ,

$ I=\frac{e}{R}=\frac{500}{20}=25 \mathrm{~A} $

Number of turns in flat circular coil, $N=100$

Radius, $r=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Rate of change of magnetic field,

$ \frac{d B}{d t}=0.1 \mathrm{~T} / \mathrm{s} $

∴ Induced emf in the coil,

$ \begin{aligned} |e| & =N \frac{d \phi}{d t}=N \cdot \frac{d}{d t} B A \quad[\because \phi=B A] \\ & =N \frac{d}{d t} B \cdot \pi r^2=N \cdot \pi r^2 \cdot \frac{d B}{d t} \\ & =100 \times \pi \times(0.1)^2 \times 0.1=\frac{\pi}{10} \mathrm{~V} \end{aligned} $

Given, number of turns per cm in long solenoid, $n=20$ turns $/ \mathrm{cm}=2000$ turns/m

$ \begin{aligned} & \text { Area of loop, } A=\frac{4}{\pi} \mathrm{~cm}^2=\frac{4}{\pi} \times 10^4 \mathrm{~m}^2 \\ & \Delta I=3-1 \neq 2 \mathrm{~A} \quad \Delta t=0.2 \mathrm{~s} \end{aligned} $

∴ Magnetic field inside the solenoid,

$ B=\mu_0 n I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Magnetic flux linked with loop kept inside the solenoid normal to it,

$ \phi=B A=\mu_0 n I A\,\,\,\,\,\,\, [from Eq. (i)]$

∴ According to Faraday's law of electromagnetic induction, magnitude of induced emf is given as

$ \begin{aligned} & |e|=\frac{d \phi}{d t}=\frac{d}{d t}\left(\mu_0 n I A\right)=\mu_0 n A \frac{d I}{d t} \\ & \Rightarrow|e|=\mu_0 n A \cdot \frac{\Delta I}{\Delta t}=4 \pi \times 10^{-7} \times 2000 \times \frac{4}{\pi} \times 10^{-4} \times \frac{2}{0.2} \\ & =3.2 \times 10^{-6} \mathrm{~V}=3.2 \mu \mathrm{~V}\left(\because 10^{-6} \mathrm{~V}=1 \mu \mathrm{~V}\right) \end{aligned} $

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

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

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

According to given circuit diagram, equivalent resistance between point P and Q.

${R_{PQ}} = (4 + 4)||(4 + 4)$

$ = {{8 \times 8} \over {8 + 8}} = 4\,\Omega $

The equivalent circuit can be drawn as,


Equivalent resistance, R_eq = 4 + 1 = 5 $\Omega$

Magnetic field, B = 5T

The side of the square loop, I = 20 cm = 0.20 m

The steady value of the current, I = 2 mA = 2 $\times$ 10^-3 A

Induced emf, e = Bv₀I

Induced current, I = e / R_eq

Substituting the values in the above equation, we get

$2 \times {10^{ - 3}} = {{5 \times {v_0} \times 0.2} \over 5}$

$\Rightarrow$ v₀ = 10^-2 m/s = 1 cm/s

$\therefore$ The value of v₀ = 1 cm/s, so that a steady current of 2 mA flows in the loop.

$\overrightarrow B $ must not be parallel to the plane of coil for non zero flux and according to lenz law if B is outward it should be decreasing for anticlockwise induced current.

$B = \left[ {{{{\mu _0}} \over {4\pi }}{I \over {b/2}} \times 2\sin 45} \right] \times 4$

$\phi = 2\sqrt 2 {{{\mu _0}} \over \pi }{I \over b} \times {a^2}$

$\therefore$ $M = {\phi \over I} = {{2\sqrt 2 {\mu _0}{a^2}} \over {\pi b}} = {{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}$

Option (a)

emf = BLV

= 1 . (2R) . 1

= 2V

$\to$ When magnet passes through centre region of solenoid, no current / Emf is induced in loop.

$\to$ While entering flux increases so negative induced emf

$\to$ While leaving flux decreases so positive induced emf.

$U = {1 \over 2}L{i^2} = 64 \Rightarrow L = 2$

${i^2}R = 640$

$R = {{640} \over {{{(8)}^2}}} = 10$

$\tau = {L \over R} = {1 \over 5} = 0.2$

As the rectangular conductor moves in field area, so flux is increasing up to x = b, then flux is generated on return journey from x = b to x = 0. The flux is shown by plot A of the graph.

As, emf, $e = - {{d\phi } \over {dt}}$, which is shown by curve B and power dissipated, P = VI which is shown by curve C.

Magnetic energy, U = ${1 \over 2}$LI$_0^2$

Given : U = 25% of U₀.

$ \Rightarrow $ ${1 \over 2}L{I^2} = {1 \over 4} \times {1 \over 2}LI_0^2$

$ \Rightarrow {I^2} = {{I_0^2} \over 4} \Rightarrow I = {{{I_0}} \over 2}$

$ \therefore $ $I = {I_0}(1 - {e^{ - t/\tau }})$

$ \Rightarrow {{{I_0}} \over 2} = {I_0}(1 - {e^{ - t/\tau }})$

$ \Rightarrow {1 \over 2} = {e^{ - t/\tau }}$

$ \Rightarrow {e^{t/\tau }} = 2$

$ \Rightarrow t = \tau \ln 2$

$ \Rightarrow t = {L \over R}\ln 2$

We know, Emf = Blv

where B, l and v are mutually perpendicular to each other. If any two are not perpendicular to each other then Emf = 0.

Here for part AB and DC, length l and velocity v is parallel to each other so, Emf by AB and DC is zero.

Given $\overrightarrow B = {B_0}\left( {{x \over a}} \right)\widehat k$

At x = 0,

$B = {B_0}\left( {{0 \over a}} \right)$

$ = 0$

$ \therefore $ For part AD, Emf = 0

In BC part,

x = d

$ \therefore $ $B = {B_0}\left( {{d \over a}} \right)$

And Emf for BC part,

$\varepsilon $ $ = Bvd$

Here B, v, d are mutually perpendicular to each other.

$ \therefore $ $\varepsilon $ $ = {B_0}\left( {{d \over a}} \right) \times {v_0} \times d$

$ = {{{B_0}{v_0}{d^2}} \over a}$

When bar slides, area of loop 1 decreases and that of loop 2 increases. Magnetic flux decreases in 1 and increases in 2. Therefore induced emf and current resist this change. As a result B should increase in 1 and decrease in 2. So, I₁ should be clockwise and I₂ anticlockwise.

$\varepsilon $_ind = (B_v) LV and B_v = B_Total sin60^o

$ \therefore $ $\varepsilon $_ind = (2.5 $\times$ 10^$-$4)(sin 60^o) $\times$ 10 $\times$ 180 $\times$ ${5 \over {18}}$

= 108.25 mV