Electromagnetic Induction

The force on a current-carrying wire is given by $\mathrm{F}=\mathrm{BI} \mathrm{I} \sin \theta$

$ \mathrm{B}=\frac{\mathrm{F}}{\mathrm{I} \cdot \mathrm{l} \sin \theta} $

Dimensions of force $[\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right]$

Dimensions of current $[\mathrm{I}]=[\mathrm{A}]$

Dimension of length $[\mathrm{l}]=[\mathrm{L}]$

Dimension of trigonometric function is $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$, so $[\sin \theta]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$

So, the dimension of magnetic field is,

$ [\mathrm{B}]=\left[\mathrm{MLT}^{-2}\right][\mathrm{A}][\mathrm{L}]=\left[\mathrm{M} \mathrm{~L}^{1-1} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right] $

$\Rightarrow $ $ [\mathrm{B}]=\left[\mathrm{M} \mathrm{~L}^0 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right] \rightarrow(\mathrm{III}) $

Magnetic flux is the product of magnetic field and area: $\Phi=\mathrm{B} \cdot \mathrm{A}$

Dimensions of magnetic induction $[\mathrm{B}]=\left[\mathrm{M} \mathrm{T}^{-2} \mathrm{~A}^{-1}\right]$

Dimensions of area $[\mathrm{A}]=\left[\mathrm{L}^2\right]$

So, the dimension of magnetic flux is,

$ [\Phi]=\left[\mathrm{M} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right] \times\left[\mathrm{L}^2\right] $

$\Rightarrow $ $[\Phi]=\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right] \rightarrow(\mathrm{IV})$

From Ampere's Law or the force between two parallel wires :

$\mathrm{F}=\frac{\mu \mathrm{I}_1 \mathrm{I}_2 \mathrm{l}}{2 \pi \mathrm{r}}$.

$\Rightarrow $$ \mu=\frac{\mathrm{F} \cdot 2 \pi \mathrm{r}}{\mathrm{I}_1 \mathrm{I}_2 \cdot \mathrm{l}} $

Dimensions of Force $[\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right]$

Dimensions of distance $[\mathrm{r}]=[\mathrm{L}]$

Dimensions of 2 and $\pi$ is $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$

Dimensions of current $[\mathrm{I}]=[\mathrm{A}]$

Dimensions of length $[\mathrm{l}]=[\mathrm{L}]$

So, the dimensions of magnetic permeability is,

$ [\mu]=\frac{\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]}{[\mathrm{A}][\mathrm{A}][\mathrm{L}]} $

$\Rightarrow $ $[\mu]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{A}^2\right][\mathrm{L}]}=\left[\mathrm{ML}^{2-1} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]$

$\Rightarrow $ $ [\mu]=\left[\mathrm{M} \mathrm{~L} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right] \rightarrow(\mathrm{I}) $

The energy stored in an inductor is $\mathrm{U}=\frac{1}{2} \mathrm{LI}^2$.

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

Dimensions of energy $[\mathrm{U}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

Dimensions of current $[\mathrm{I}]=[\mathrm{A}]$

So, the dimensions of self-inductance is,

$[\mathrm{L}]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{A}^2\right]}$

$\Rightarrow $ $ [\mathrm{L}]=\left[\mathrm{M} \mathrm{~L}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right] \rightarrow(\mathrm{II}) $

Hence, the correct match is A-III, B-IV, C-I, D-II. Therefore, the correct option is (D).

According to Faraday's Law of Electromagnetic Induction, the induced EMF (E) is equal to the rate of change of magnetic flux $(\phi)$ through the loop.

Initially circular loop is with radius $\mathrm{r}=7 \mathrm{~cm}=0.07 \mathrm{~m}$.

Finally a square loop formed from the same wire.

Magnetic field $\mathrm{B}=0.2 \mathrm{~T}$ (constant and perpendicular).

Time interval $=\Delta \mathrm{t}=0.5 \mathrm{~s}$.

Initial Area ( $\mathrm{A}_1$ )

$ \mathrm{A}_1=\pi \mathrm{r}^2=\pi \times(0.07)^2=0.0154 \mathrm{~m}^2 $

The perimeter of the square must be equal to the circumference of the circle because the wire length is constant.

$ 2 \pi \mathrm{r}=4 \mathrm{a} \Rightarrow \mathrm{a}=\frac{2 \pi \mathrm{r}}{4}=\frac{\pi \mathrm{r}}{2}=\frac{\pi \times 0.07}{2} \mathrm{~m}=0.11 \mathrm{~m} $

So, final area ( $\mathrm{A}_2$ )

$ a^2=(0.11)^2=0.0121 m^2 . $

Since the field is perpendicular, $\phi=\mathrm{B} \cdot \mathrm{A}$.

So, the change in magnetic flux is,

$ \Delta \phi=\mathrm{B}\left(\mathrm{~A}_2-\mathrm{A}_1\right) $

$\Rightarrow $ $\Delta \phi=0.2 \times(0.0121-0.0154)$

$\Rightarrow $ $\Delta \phi=0.2 \times(-0.0033)=-0.00066 \mathrm{~Wb}$

The magnitude of the induced EMF is :

$ |E|=\left|\frac{\Delta \phi}{\Delta t}\right| $

$\Rightarrow $ $ |\mathrm{E}|=\frac{0.00066}{0.5}=0.00132 \mathrm{~V}=0.00132 \times 1000 \mathrm{mV}=1.32 \mathrm{mV} $

Therefore, the induced EMF is 1.32 mV .

Hence, the correct option is (C).

For a long solenoid, the magnetic field B inside is uniform and given by :

$ \mathrm{B}=\mu_0 \mathrm{nI} $

Where :

• $\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{~m} / \mathrm{A}$

• $\mathrm{n}=500$ turns/unit length.

• $\mathrm{I}=10 \sin (1000 \mathrm{t}) \mathrm{A}$

The loop is inside the solenoid, so it experiences this magnetic field. The flux through the circular loop of radius $\mathrm{r}=1 \mathrm{~cm}=0.01 \mathrm{~m}$ is :

$ \Phi=\mathrm{B} \cdot \mathrm{~A}=\left(\mu_0 \mathrm{nI}\right) \cdot\left(\pi \mathrm{r}^2\right) $

$\Rightarrow $ $\Phi=\mu_0 \mathrm{n}\left(\pi \mathrm{r}^2\right) \cdot 10 \sin (1000 \mathrm{t})$

According to Faraday's Law, the induced EMF is $\varepsilon=-\frac{\mathrm{d} \Phi}{\mathrm{dt}}$.

Since the loop is already 10 cm inside a 100 cm long solenoid, it is in the region where the magnetic field is uniform. When the loop is inside the magnetic field then its motion along axis of solenoid will not cause any change in effective magnetic flux through loop because loop is co-axial with the solenoid.

$ \varepsilon=\frac{d}{d t}\left[\mu_0 n \pi r^2 \cdot 10 \sin (1000 t)\right] $

$\Rightarrow $$ \varepsilon=\mu_0 \mathrm{n} \pi \mathrm{r}^2 \cdot 10 \cdot 1000 \cos (1000 \mathrm{t}) $

The induced current i is $\frac{\varepsilon}{\mathrm{R}}$ :

$ \mathrm{i}=\frac{10000 \mu_0 \mathrm{n} \pi \mathrm{r}^2}{\mathrm{R}} \cos (1000 \mathrm{t}) $

Peak induced current $\mathrm{i}_0=\frac{10^4 \times\left(4 \pi \times 10^{-7}\right) \times 500 \times \pi \times(0.01)^2}{10} \mathrm{~A}$

$\Rightarrow $ $ \mathrm{i}_0 \approx 1.97 \times^{-4} \mathrm{~A}=197 \times 10^{-6} \mathrm{~A}=197 \mu \mathrm{~A} $

RMS current of a sinusoidal current is given as,

$ \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_0}{\sqrt{2}}=\frac{197}{\sqrt{2}} \mu \mathrm{~A}=\frac{\alpha}{\sqrt{2}} \mu \mathrm{~A} $

$\Rightarrow \alpha=197$

The wire falls from rest (initial velocity $u=0$ ) under gravity.

Using the third equation of motion its velocity after falling a distance $h=200 \mathrm{~m}$ is,

$ v^2=u^2+2 g h $

$\Rightarrow $ $v^2=0^2+2(10)(200)$

$\Rightarrow $ $\mathrm{v}^2=4000$

$\Rightarrow $ $\mathrm{v}=\sqrt{4000} \mathrm{~m} / \mathrm{s}$

$\Rightarrow $ $ \mathrm{v}=20 \sqrt{10} \mathrm{~m} / \mathrm{s} $

The formula for motional EMF induced in a conductor moving perpendicular to a magnetic field is :

$ \mathrm{E}=\mathrm{Bvl} $

Where $B$ is magnetic field strength, $v$ is the velocity of the conductor and $l$ is the length of the conductor

$ E=\left(5 \times 10^{-5}\right) \times(20 \sqrt{10}) \times(20) $

$\Rightarrow $ $E=2000 \times 10^{-5} \times \sqrt{10}$

$\Rightarrow $ $E=2 \times 10^{-2} \times \sqrt{10}$ Volts

$\Rightarrow $ $E=\left(2 \times 10^{-2} \sqrt{10}\right) \times 10^3 \mathrm{mV}$

$\Rightarrow $ $\mathrm{E}=20 \sqrt{10} \mathrm{mV}$

The ammeter is placed in the main line to measure the total current drawn from the battery.

An inductor resists any sudden change in the current flowing through it due to self-induction (Lenz's Law). The induced EMF opposes the applied voltage.

Before the switch K is turned ON , the current in the circuit is zero. At the exact moment the switch is turned ON $(t=0)$, the inductors will strongly oppose the sudden increase in current.

Therefore, at $\mathrm{t}=0$, the current through any branch containing an inductor is exactly zero. Effectively, the inductors act as open circuits.

Because the inductors act as open circuits the moment the switch is closed, no current flows through those two branches.

So, the initial current (I) drawn from the battery using Ohm's Law is:

$ I=\frac{V}{R} \Rightarrow I=\frac{9}{9} A=1 A $

Therefore, the total reading of the ammeter at the exact moment the switch is turned ON is 1 A . Hence, option (C) is correct.

As the rod falls under gravity with a downward velocity v , it cuts through the perpendicular magnetic field lines. This changing magnetic flux through the closed loop XPQY induces an EMF according to Faraday's Law of Induction.

Let the distance from the top resistor R to the rod be y.

Even though the rod has length $L$, only the length $l$ between the vertical rails completes the electrical circuit. The overhanging parts do not contribute to the closed loop's area.

So, the area of the rectangular loop formed by the rails, the resistor, and the active part of the rod is $\mathrm{A}=\mathrm{l} \times \mathrm{y}$.

The magnetic flux $\Phi$ passing through this area is :

$ \Phi=\mathrm{B} \cdot \mathrm{~A}=\mathrm{B} \cdot \mathrm{l} \cdot \mathrm{y} $

According to Faraday's Law, the magnitude of the induced EMF (e) is the rate of change of magnetic flux :

$ e=\left|\frac{d \Phi}{d t}\right|=\left|\frac{d(B \cdot l \cdot y)}{d t}\right| $

$\Rightarrow $ $ e=B l\left|\frac{d y}{d t}\right| $

The rate of change of position $y$ with respect to time is simply the velocity $v$ of the rod:

$ \mathrm{e}=\mathrm{Blv} $

The closed loop has a total resistance R. According to Ohm's Law, the induced current I flowing through the circuit is:

$ I=\frac{e}{R}=\frac{B l v}{R} $

When a current-carrying conductor is placed in a magnetic field, it experiences a magnetic force. The formula for the force on a straight segment of wire is $\overrightarrow{\mathrm{F}}_{\mathrm{m}}=\mathrm{I}\left(\overrightarrow{\mathrm{l}}_{\mathrm{eff}} \times \overrightarrow{\mathrm{B}}\right)$.

The current I only flows through the part of the rod between the rails (length l). The overhanging parts carry no current because they are open circuits.

Therefore, the effective length is l , not L .

Since the length vector is perpendicular to the magnetic field, the magnitude of the force is $\mathrm{F}_{\mathrm{m}}=\mathrm{IlB}$.

$ \mathrm{F}_{\mathrm{m}}=\left(\frac{\mathrm{Blv}}{\mathrm{R}}\right) \mathrm{lB} $

$\Rightarrow $ $ F_m=\frac{B^2 l^2 v}{R} $

According to Lenz's Law, the direction of the induced current will be such that the resulting magnetic force opposes the change that caused it. Since the rod is falling down (increasing the area), the magnetic force will act upwards to oppose the downward motion.

The net downward force is :

$\mathrm{F}_{\mathrm{net}}=\mathrm{mg}-\mathrm{F}_{\mathrm{m}}$

$\Rightarrow $ $ \mathrm{ma}=\mathrm{mg}-\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{v}}{\mathrm{R}} $

When the rod is first released from rest $(\mathrm{v}=0)$, the upward magnetic force is zero, and the rod accelerates downwards at g . As its velocity v increases, the upward magnetic force increases.

Eventually, the upward magnetic force becomes exactly equal to the downward weight. At this point, the net force becomes zero, and the acceleration (a) becomes zero. The rod then continues to fall at a constant, maximum velocity known as the terminal velocity $\left(\mathrm{v}_{\mathrm{T}}\right)$.

At terminal velocity $\left(\mathrm{v}=\mathrm{v}_{\mathrm{T}}\right), \mathrm{a}=0$ :

$ 0=\mathrm{mg}-\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{v}_{\mathrm{T}}}{\mathrm{R}} $

$\Rightarrow $ $\mathrm{mg}=\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{v}_{\mathrm{T}}}{\mathrm{R}}$

$\Rightarrow $ $ \mathrm{v}_{\mathrm{T}}=\frac{\mathrm{mgR}}{\mathrm{~B}^2 \mathrm{l}^2} $

Therefore, the terminal speed of rod is $\frac{m g R}{B^2 1^2}$. Hence, the correct option is (A).

Looking at the coils along their common axis from the right side looking left).

Coil $\mathrm{C}_1$ carries current in the anti-clockwise ( ACW ) direction. According to the Right-Hand Grip Rule, an ACW current produces a magnetic field $\left(\overrightarrow{\mathrm{B}}_1\right)$ pointing towards the observer.

Let's define this direction towards the observer as positive flux.

Coil $\mathrm{C}_3$ : Carries current in the clockwise (CW) direction. An ACW current produces a magnetic field ( $\overrightarrow{\mathrm{B}}_3$ ) pointing away from the observer.

Let's define this direction pointing away from the observer as negative flux.

Coil $C_2$ is exactly in the middle. Because $C_1$ and $C_3$ are identical and carry the same magnitude of current $I$, the strength of their magnetic fields at the midpoint will be identical.

Since $\vec{B}$ points towards us and $\vec{B}_3$ points away from us with equal magnitudes, they cancel each other out.

Initial Net Magnetic Flux through $\mathrm{C}_2$ is $\Phi_{\text {initial }}=0$

A clockwise current in $\mathrm{C}_2$ would create its own induced magnetic field $\left(\overrightarrow{\mathrm{B}}_{\text {induced }}\right)$ pointing away from the observer i.e., negative direction.

Lenz's Law states that an induced current will always flow in a direction that opposes the change in magnetic flux that caused it.

Since the induced field points in the negative direction, it must be resisting a change in the positive direction.

Therefore, the external magnetic flux through $\mathrm{C}_2$ must be increasing in the positive direction (towards the observer).

To get a net increase in positive flux, we have two ways :

• Increase the positive flux from $\mathrm{C}_1$ by moving the source of the positive flux $\left(\mathrm{C}_1\right)$ closer to $\mathrm{C}_2$.

• Decrease the negative flux from $\mathrm{C}_3$ by moving the source of the negative flux $\left(\mathrm{C}_3\right)$ further away from $\mathrm{C}_2$.

Therefore, $C_1$ moves towards $C_2$ (increases positive flux) and $C_3$ moves away from $C_2$ (decreases negative flux). This results net positive flux change required to induce a clockwise current.

Hence, the correct option is (B).

According to Faraday's Law of Electromagnetic Induction, the induced $\operatorname{emf}(\varepsilon)$ is the negative rate of change of magnetic flux $(\phi)$ through the loop :

$ \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}} $

When the field is perpendicular to the plane of the loop then magnetic flux is

$ \phi=\mathrm{B} \cdot \mathrm{~A} $

The area of the loop is $\mathrm{A}=1.0 \mathrm{~m}^2$

The magnetic field in the region is $B=\sin (100 t)$

So, the magnetic flux is,

$ \phi=1.0 \sin (100 t)=\sin (100 t) $

Now on differentiating the flux with respect to time ( $t$ ) we get the induced emf :

$ \varepsilon=-\frac{d}{d t}[\sin (100 t)] $

$\Rightarrow $ $ \varepsilon=-100 \cos (100 t) \text { Volts } $

The instantaneous power $(\mathrm{P})$ dissipated as heat in a resistor is:

$ \mathrm{P}=\frac{\varepsilon^2}{\mathrm{R}} $

Given resistance of the loop is $\mathrm{R}=100 \Omega$ :

$ P=\frac{[-100 \cos (100 t)]^2}{100} $

$\Rightarrow $ $ \mathrm{P}=\frac{10000 \cos ^2(100 \mathrm{t})}{100}=100 \cos ^2(100 \mathrm{t}) \text { Watts } $

The average thermal energy dissipated in one period is equivalent to finding the average power over one full cycle ( T ) in time T .

The time period $(T)$ for the function $\sin (100 t)$ is,

$ \mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{100}=\frac{\pi}{50} \text { seconds } $

The total energy (E) dissipated in one period is :

$ \mathrm{E}=\int_0^{\mathrm{T}} \mathrm{Pdt}=\int_0^{\mathrm{T}}\left(100 \cos ^2(100 \mathrm{t})\right) \mathrm{dt} $

$\Rightarrow $ $\mathrm{E}=100 \int_0^{\mathrm{T}} \cos ^2(100 \mathrm{t}) \mathrm{dt}=100 \int_0^{\mathrm{T}} \frac{1+\cos (200 \mathrm{t})}{2} \mathrm{dt}$

$\Rightarrow $ $E=50 \int_0^T 1+\cos (200 t) d t$

$\Rightarrow $ $E=50\left[t+\frac{\sin (200 t)}{200}\right]_0^{\pi / 50}$

$\Rightarrow $ $\mathrm{E}=50\left[\left(\frac{\pi}{50}+\frac{\sin \left(200 \times \frac{\pi}{50}\right)}{200}\right)-(0+0)\right]$

$\Rightarrow $ $E=50\left[\frac{\pi}{50}+\frac{\sin (4 \pi)}{200}\right]$

$\Rightarrow $ $\mathrm{E}=50 \times \frac{\pi}{50}=\pi$ Joules

When a conductor moves through a uniform magnetic field, it experiences motional EMF. For a rod of length l moving with velocity v perpendicular to a magnetic field B , the induced emf is :

$ \varepsilon=\mathrm{Bvl} $

The given values are,

• $\mathrm{B}=0.10 \mathrm{~T}$

• $\mathrm{v}=1.5 \mathrm{~m} / \mathrm{s}$

• $\mathrm{l}=1 \mathrm{~m}$

Putting the values,

$ \varepsilon=(0.10) \times(1.5) \times(1)=0.15 \text { Volts } $

Using Ohm's Law, where $R$ is the total resistance of the circuit:

$ I=\frac{\varepsilon}{R} $

Given $\mathrm{R}=2 \Omega$ :

$ I=\frac{0.15}{2}=0.075 \text { Amperes } $

A current-carrying wire in a magnetic field experiences a force given by:

$ F_m=I l B \sin (\theta) $

Since the rod is perpendicular to the field, $\theta=90^{\circ}$,

$ \Rightarrow \mathrm{F}_{\mathrm{m}}=\mathrm{IlB} $

According to Lenz's Law, this magnetic force will act in a direction opposite to the motion (towards the left).

$ \mathrm{F}_{\mathrm{m}}=(0.075) \times(1) \times(0.10) $

$\Rightarrow $ $ \mathrm{F}_{\mathrm{m}}=0.0075 \mathrm{~N}=7.5 \times 10^{-3} \mathrm{~N} $

Then the net force to move the rod with constant speed must be zero. Therefore, the external force ( $\mathrm{F}_{\text {ext }}$ ) must be equal in magnitude to the magnetic force:

$ \mathrm{F}_{\mathrm{ext}}=\mathrm{F}_{\mathrm{m}}=7.5 \times 10^{-3} \mathrm{~N} $

Therefore, the external force required to maintain the constant velocity of $1.5 \mathrm{~m} / \mathrm{s}$ against the magnetic resistance is $7.5 \times 10^{-3} \mathrm{~N}$.

Hence, the correct option is (B).

According to Faraday's Law, a change in the magnetic flux $(\Phi)$ through a coil induces an electromotive force (e.m.f.) in that coil. The magnitude of the induced e.m.f. ( $\varepsilon$ ) is given by:

$ \varepsilon=N \frac{d \Phi}{d t} $

Where N is the total number of turns. For a solenoid, the magnetic flux $\Phi$ through each turn is $\Phi=\mathrm{B} \cdot \mathrm{A}$, where $B$ is the magnetic field inside the solenoid and $A$ is the cross-sectional area.

The magnetic field $B$ inside an ideal solenoid is:

$ \mathrm{B}=\mu_0 \mathrm{nI} $

Where $\mu_0$ is the permeability of free space $\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \frac{\mathrm{~m}}{\mathrm{~A}}\right)$, n is the number of turns per unit length and I is the current.

The total flux linkage is $\Psi=N \Phi=(n l)(B A)=(n l)\left(\mu_0 n I A\right)=\mu_0 n^2 l A I$.

Since $\varepsilon=\frac{\mathrm{d} \Psi}{\mathrm{dt}}$,

$ \varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}} $

Where the self-inductance L is $\mathrm{L}=\mu_0 \mathrm{n}^2$ lA

The length of the solenoid is $\mathrm{l}=30 \mathrm{~cm}=0.3 \mathrm{~m}$

Turns per $\mathrm{cm}=10$, so turns per meter will be $\mathrm{n}=10 \times 100=1000$ turns $/ \mathrm{m}$.

The area of cross-section is $\mathrm{A}=5 \mathrm{~cm}^2=5 \times 10^{-4} \mathrm{~m}^2$.

So,

$L=\left(4 \times 3.14 \times 10^{-7}\right) \times(1000)^2 \times 0.3 \times\left(5 \times 10^{-4}\right)$

$\Rightarrow $ $\mathrm{L}=12.56 \times 10^{-7} \times 10^6 \times 0.3 \times 5 \times 10^{-4}$

$\Rightarrow $ $ \mathrm{L}=12.56 \times 1.5 \times 10^{-5}=18.84 \times 10^{-5} \mathrm{H} $

The change in current is $\Delta \mathrm{I}=4 \mathrm{~A}-2 \mathrm{~A}=2 \mathrm{~A}$.

The time interval ( $\Delta \mathrm{t}$ ) is 3.14 s.

$ \varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}} $

$\Rightarrow $ $\varepsilon=\left(18.84 \times 10^{-5}\right) \times \frac{2}{3.14}$

$\Rightarrow $ $\varepsilon=6 \times 2 \times 10^{-5}$

$\Rightarrow $ $ \varepsilon=12 \times 10^{-5} \mathrm{~V}=\alpha \times 10^{-5} \mathrm{~V} \Rightarrow \alpha=12 $

Therefore, the value of $\alpha$ is 12 .

Hence, the correct option is (B).

Magnetic flux $\phi$ through a loop is $\phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos \theta$.

According to Faraday's Law, the induced emf $\varepsilon$ is the negative rate of change of flux:

$ \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA} \cos \theta) $

Since area (A) and angle ( $\theta$ ) are constant.

$ \varepsilon=-\mathrm{A} \cos \theta \frac{\mathrm{~dB}}{\mathrm{dt}} $

The length of side of square is $\mathrm{s}=2 \mathrm{~cm}=0.02 \mathrm{~m}$.

Then the area of loop is,

$ A=s^2=(0.02)^2=4 \times 10^{-4} \mathrm{~m}^2 $

The angle made by magnetic field with the normal to the plane of loop is $\theta=60^{\circ}$

The time varying magnetic field is $B=0.4 \sin (300 t)$.

The rate of change of magnetic field is

$ \frac{\mathrm{dB}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(0.4 \sin (300 \mathrm{t}))=0.4 \times 300 \cos (300 \mathrm{t})=120 \cos (300 \mathrm{t}) $

Substituting into the emf equation :

$ \varepsilon=-\left(4 \times 10^{-4}\right) \cos \left(60^{\circ}\right)[120 \cos (300 \mathrm{t})] $

$\Rightarrow $ $ \varepsilon=-\left(4 \times 10^{-4}\right) \times 0.5 \times 120 \cos (300 t)=-0.024 \cos (300 t) V $

The maximum induced emf $\left(\varepsilon_{\text {max }}\right)$ occurs when $\cos (300 t)=1$ :

$ \varepsilon_{\max }=0.024 \mathrm{~V}=24 \mathrm{mV} $

Therefore, the magnitude of maximum induced emf is 24 mV .

Thus, the correct option is (D).

When a battery is connected to an inductor $(\mathrm{L})$ and a resistor $(\mathrm{R})$ in series, the total voltage $(\mathrm{E})$ provided by the battery is shared between the resistance and the inductance according to Kirchhoff's Voltage Law:

$ E=V_R+V_L $

Where, E is the E.M.F. of the battery. $\mathrm{V}_{\mathrm{R}}$ is the voltage across the resistance, calculated as $\mathrm{V}_{\mathrm{R}}=\mathrm{iR} . \mathrm{V}_{\mathrm{L}}$ is the instantaneous voltage across the inductor.

$ V_L=E-i R $

The emf of the battery is $\mathrm{E}=1.0 \mathrm{~V}$ and connected resistance is $\mathrm{R}=100 \Omega$.

Case 1 when the current passing through inductor is $\mathrm{i}_1=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A}$.

$ V_{L 1}=1.0-\left(2 \times 10^{-3} \times 100\right)=1.0-0.2=0.8 \mathrm{~V} $

Case 2 when the current passing through inductor is $\mathrm{i}_2=4 \mathrm{~mA}=4 \times 10^{-3} \mathrm{~A}$.

$ V_{L 2}=1.0-\left(4 \times 10^{-3} \times 100\right)=1.0-0.4=0.6 \mathrm{~V} $

So, the ratio of the instantaneous voltages is:

$ \frac{V_{L 1}}{V_{L 2}}=\frac{0.8}{0.6}=\frac{8}{6}=\frac{4}{3} $

Therefore, the correct option is (A).

For a rod rotating about one end with angular velocity $\omega$, the small element at distance $r$ from the origin has linear speed

$ v=\omega r $

If the magnetic field at distance $r$ is

$ B(r)=B_0 e^{-\lambda r} $

then the small emf induced across a small length $dr$ is

$ d\mathcal{E}=B(r)\,v\,dr $

So,

$ d\mathcal{E}=B_0 e^{-\lambda r}\,\omega r\,dr $

Hence total emf across the rod is

$ \mathcal{E}=\int_0^L B_0 \omega r e^{-\lambda r}\,dr $

Take constants outside:

$ \mathcal{E}=B_0\omega \int_0^L r e^{-\lambda r}\,dr $

Now evaluate

$ I=\int r e^{-\lambda r}\,dr $

Using integration by parts:

Let

$ u=r,\qquad dv=e^{-\lambda r}dr $

Then

$ du=dr,\qquad v=-\frac{1}{\lambda}e^{-\lambda r} $

So,

$ I=uv-\int v\,du $

$ I=-\frac{r}{\lambda}e^{-\lambda r}+\frac{1}{\lambda}\int e^{-\lambda r}dr $

$ I=-\frac{r}{\lambda}e^{-\lambda r}-\frac{1}{\lambda^2}e^{-\lambda r} $

Thus,

$ \int_0^L r e^{-\lambda r}\,dr = \left[-e^{-\lambda r}\left(\frac{r}{\lambda}+\frac{1}{\lambda^2}\right)\right]_0^L $

Substitute limits:

$ = -e^{-\lambda L}\left(\frac{L}{\lambda}+\frac{1}{\lambda^2}\right)-\left[-1\cdot\frac{1}{\lambda^2}\right] $

$ = \frac{1}{\lambda^2}-e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right) $

Therefore,

$ \mathcal{E}=B_0 \omega\left[\frac{1}{\lambda^2}-e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right] $

So the correct option is:

$ \boxed{\text{Option A}} $

The outer long solenoid has length $d$, area $S$, and number of turns per unit length $N$

It connected to a capacitor C, forming an LC circuit.

The inner short solenoid has length $\frac{\mathrm{d}}{2}$, area $\frac{\mathrm{S}}{2}$, and number of turns per unit length 2 N is placed completely inside it and short-circuited.

For a long solenoid where end-effects are negligible ( $\mathrm{d}^2 \gg \mathrm{~S}$ )

The magnetic field inside it when carrying a current $\mathrm{I}_1$ is :

$ \mathrm{B}_1=\mu_0 \mathrm{NI}_1 $

The total number of turns in the outer coil is $\mathrm{N}_1=\mathrm{N} \cdot \mathrm{d}$.

The total magnetic flux $\Phi_1$ linking the outer coil itself is :

$ \Phi_1=N_1\left(B_1 S\right)=(N d)\left(\mu_0 N I_1 S\right)=\mu_0 N^2 S d I_1 $

The self-inductance is $\mathrm{L}=\Phi_1 \mathrm{I}_1$ :

$ \mathrm{L}=\mu_0 \mathrm{~N}^2 \mathrm{Sd} $

The length of the inner coil is $\mathrm{d}_2=\frac{\mathrm{d}}{2}$ and cross-sectional area is $\mathrm{S}_2=\frac{\mathrm{S}}{2}$.

Its number of turns per unit length is $\mathrm{N}_2=2 \mathrm{~N}$.

The total number of turns in the inner coil is:

$ \mathrm{N}_{\text {total }, 2}=\mathrm{N}_2 \cdot \mathrm{~d}_2=(2 \mathrm{~N})\left(\frac{\mathrm{d}}{2}\right)=\mathrm{Nd} $

Using the self-inductance of a solenoid,

$ \mathrm{L}_2=\mu_0 \mathrm{~N}_2^2 \mathrm{~S}_2 \mathrm{~d}_2 $

$\Rightarrow $ $L_2=\mu_0(2 N)^2\left(\frac{S}{2}\right)\left(\frac{d}{2}\right)=\mu_0\left(4 N^2\right) \frac{S d}{4}=\mu_0 N^2 S d$

$\Rightarrow $ $ \mathrm{L}_2=\mathrm{L} $

When a current $\mathrm{I}_1$ flows through the outer coil, it establishes a uniform magnetic field $\mathrm{B}_1=\mu_0 \mathrm{~N} \mathrm{I}_1$ inside its core. Since the inner coil is completely inside the outer coil, it experiences this exact magnetic field.

The total flux $\Phi_{21}$ linking the inner coil due to the outer coil's current is :

$ \Phi_{21}=N_{\text {total }, 2}\left(B_1 S_2\right) $

$ \Phi_{21}=(\mathrm{Nd})\left(\mu_0 \mathrm{NI}_1\right)\left(\frac{\mathrm{S}}{2}\right)=\frac{1}{2} \mu_0 \mathrm{~N}^2 \mathrm{~S} \mathrm{~d} \mathrm{I}_1 $

The mutual inductance M is given by $\mathrm{M}=\Phi_{21} \mathrm{I}_1$ :

$ M=\frac{1}{2} \mu_0 N^2 S d=\frac{L}{2} $

Let q be the charge on the capacitor C at any instant, and $\mathrm{I}_1=-\frac{\mathrm{dq}}{\mathrm{dt}}$ be the current flowing through the outer loop containing the capacitor.

Let $\mathrm{I}_2$ be the induced current flowing through the short-circuited inner loop.

Applying Kirchhoff’s Voltage Law (KVL) to inner short-circuited loop 2 :

The total induced EMF in the inner loop must be zero because it has no ohmic resistance or external voltage sources (R=0) :

$-L_2 \frac{dI_2}{dt} - M \frac{dI_1}{dt} = 0$

Using $L_2 = L$ and $M = \frac{L}{2}$:

$-L \frac{dI_2}{dt} - \frac{L}{2} \frac{dI_1}{dt} = 0 \rightarrow \frac{dI_2}{dt} = -\frac{1}{2} \frac{dI_1}{dt}$

Assuming the system starts from rest where initial currents are zero:

$I_2 = -\frac{1}{2} I_1$

Applying KVL to the loop 1

$\frac{q}{C} - L \frac{dI_1}{dt} - M \frac{dI_2}{dt} = 0$

Substituting $M = \frac{L}{2}$ and $\frac{dI_2}{dt} = -\frac{1}{2} \frac{dI_1}{dt}$

$\frac{q}{C} - L \frac{dI_1}{dt} - \left(\frac{L}{2}\right)\left(-\frac{1}{2} \frac{dI_1}{dt}\right) = 0$

$\frac{q}{C} - L \frac{dI_1}{dt} + \frac{L}{4} \frac{dI_1}{dt} = 0$

$\frac{q}{C} - \frac{3}{4} L \frac{dI_1}{dt} = 0$

Since $I_1 = -\frac{dq}{dt} \rightarrow \frac{dI_1}{dt} = -\frac{d^2 q}{dt^2}$:

$\frac{q}{C} - \frac{3}{4} L \left(-\frac{d^2 q}{dt^2}\right) = 0$

$\left(\frac{3}{4} L\right) \frac{d^2 q}{dt^2} + \left(\frac{1}{C}\right) q = 0$

$\frac{d^2 q}{dt^2} + \left(\frac{4}{3LC}\right) q = 0$

On comparing with standard differential equation for simple harmonic motion,

$\frac{d^2 q}{dt^2} + \omega_0^2 q = 0,$

where $\omega_0$ is the angular resonant frequency of the system:

$\omega_0^2 = \frac{4}{3LC}$

$\omega_0 = \sqrt{\frac{4}{3LC}} = \frac{2}{\sqrt{3LC}}$

Therefore, the correct option is (C).

When a rigid conducting loop rotates with a constant angular velocity $\omega=\frac{2 \pi}{\mathrm{~T}}$ around a fixed axis through O perpendicular to a magnetic field boundary ( $\mathrm{x}=0$ ), an electromotive force (EMF) is induced across any segment cutting through the flux zone.

According to Faraday's law of induction, the induced EMF $\varepsilon$ is given by the rate of change of magnetic flux $\Phi_{\mathrm{B}}$ through the loop:

$ \varepsilon=-\frac{\mathrm{d} \Phi_{\mathrm{B}}}{\mathrm{dt}} $

Since the magnetic field B is uniform and perpendicular to the loop area $\left(\mathrm{A}_{\mathrm{in}}\right)$ that has entered the $\mathrm{x}>0$ region:

$ \Phi_{\mathrm{B}}=\mathrm{B} \cdot \mathrm{~A}_{\mathrm{in}} \Rightarrow \varepsilon=-\mathrm{Bd} \mathrm{~A}_{\mathrm{in}} \mathrm{dt} $

The induced current $\mathrm{i}(\mathrm{t})$ passing through a loop with net resistance R is:

$ i(t)=\frac{\varepsilon}{R}=-\frac{B}{R} \frac{d A_{i n}}{d t} $

Therefore, the magnitude and sign of the current $\mathrm{i}(\mathrm{t})$ at any instant depend entirely on how the area inside the magnetic field region changes with time $\left(\frac{\mathrm{d} \mathrm{A}_{\text {in }}}{\mathrm{dt}}\right)$.

It rotates clockwise into the $\mathrm{x}>0$ field region.

As the straight edge enters the field, the sector area inside the field increases linearly with the angle $\theta = \omega t$:

$A_{\mathrm{in}}(t)=\frac{1}{2}r^2\theta=\frac{1}{2}r^2\omega t$

$\frac{dA_{\mathrm{in}}}{dt}=\frac{1}{2}r^2\omega=\text{constant}$

This yields a constant, uniform induced current value ($i=\text{constant}$) during the entry phase.

If the angular speed is constant $\omega$, time taken to enter the magnetic field is $t_1=\frac{\pi}{\omega}=\frac{T}{2}$

After that it starts moving out of the magnetic field region with same angular speed $\omega$

$\frac{dA_{\mathrm{in}}}{dt}=-\frac{1}{2}r^2\omega=\text{constant}$

So, the direction of induced current gets reversed.

If the angular speed is constant $\omega$, time taken to exit the magnetic field is $t_2=\frac{\pi}{\omega}=\frac{T}{2}$

So, the induced current will be constant for first half (positive) and constant for another half (negative).

(P)→(3)

Since the sector angle is $60^{\circ}$, it takes a time interval $\Delta t_1=\frac{T}{6}$ to completely enter the field. Once fully inside the region $\mathrm{x}>0$, the area immersed in the magnetic field remains constant for a while, meaning $\frac{\mathrm{dA}_{\text {in }}}{\mathrm{dt}}=0$, so the induced current drops sharply to zero.

The time interval till second sector is about to enter is $\Delta t_2=\frac{T}{6}$. So from $t=0$ to $t=\frac{T}{6}$ the current is constant and positive and from $t=\frac{T}{6}$ to $t=\frac{T}{6}+\Delta t_2=\frac{T}{6}+\frac{T}{6}=\frac{T}{3}$ the current is zero.

When the second sector starts to enter the magnetic field the first sector is still completely inside the magnetic field.

It takes a time interval $\Delta t_3=\frac{T}{6}$ to completely enter the field. Once fully inside the region $x>0$, the area immersed in the magnetic field remains constant for a while, meaning $\frac{\mathrm{dA}_{\text {in }}}{\mathrm{dt}}=0$, so the induced current drops sharply to zero.

So from $t=\frac{T}{3}$ to $t=\frac{T}{3}+\Delta t_3=\frac{T}{3}+\frac{T}{6}=\frac{2 T}{3}$ the current is constant and positive.

When the first sector starts leaving the field region, the area decreases at a constant rate, yielding a constant negative induced current value.

So, the cycle repeats.

(Q)→2

Since the sector angle is $60^{\circ}$, it takes a time interval $\Delta t_1=\frac{\pi / 3}{\omega}=\frac{\pi}{3 \omega}$ to completely enter the field. Once fully inside the region $\mathrm{x}>0$, the area immersed in the magnetic field remains constant for a while, meaning $\frac{\mathrm{dA} \mathrm{A}_{\text {in }}}{\mathrm{dt}}=$ 0 , so the induced current drops sharply to zero.

So from $t=0$ to $t=\frac{T}{6}$ the current is constant and positive.

The angular displacement before start leaving the magnetic field is $\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

So, the time till when the sector is completely inside the magnetic field is $\Delta \mathrm{t}_2=\frac{2 \pi / 3}{\omega}=\frac{2 \pi}{3 \omega}=2 \Delta \mathrm{t}_1$

Thus, from $\mathrm{t}=\frac{\mathrm{T}}{6}$ to $\mathrm{t}=\frac{\mathrm{T}}{6}+\Delta \mathrm{t}_2=\frac{\mathrm{T}}{6}+\frac{\mathrm{T}}{3}=\frac{2 \mathrm{~T}}{3}$

Now, the sector starts leaving the field region, the area decreases at a constant rate, yielding a constant negative induced current value for the time interval $\Delta t_3=\frac{\pi}{3 \omega}$

And then cycle repeats.

(R)→(1)

As the first sector is inside the magnetic field completely and another sector is out of magnetic field.

The angular displacement before start leaving the magnetic field is $\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

So, the time till when the sector is completely inside the magnetic field is $\Delta \mathrm{t}_1=\frac{2 \pi / 3}{\omega}=\frac{2 \pi}{3 \omega}$

Thus, from $t=0$ to $t=\frac{2 \pi}{3 \omega}$.

Now, the first sector is leaving the magnetic field with constant angular speed $\omega$ and another sector is entering the magnetic field with same angular speed.

So, the net rate of area inside the magnetic field is zero $\frac{\mathrm{dA}_{\text {in }}}{\mathrm{dt}}=0$, thus the current in loop is zero for another $\mathrm{t}= \frac{2 \pi}{3}$ time interval and the cycle repeats.

(S)→4

Therefore, the correct match is $\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 2, \mathrm{R} \rightarrow 1, \mathrm{~S} \rightarrow 4$.

So, the correct option is (C)

$\begin{aligned} &\begin{aligned} \phi & =\mathrm{BAN} \cdot \cos (\omega \mathrm{t}) \\ \varepsilon & =\frac{-\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{BA} \omega \mathrm{~N} \cdot \sin (\omega \mathrm{t}) \end{aligned}\\ &\text { When } B \text { is parallel to plane, } \underline{\underline{\omega}} \mathrm{t}=\frac{\pi}{2}\\ &\Rightarrow \phi=0, \varepsilon=\mathrm{BA} \omega \mathrm{~N} \end{aligned}$

$\begin{aligned} & \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \\ & \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}_1}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}}{\mathrm{dt}} \end{aligned}$

To determine the potential difference between the center and the rim of the disc (often called a Faraday disc or unipolar generator), we use the formula for motional EMF in a rotating disc:

$ V = \frac{1}{2} B \omega R^2 $

where:

$B$ is the magnetic field strength,

$\omega$ is the angular velocity,

$R$ is the radius of the disc.

Given:

$B = 0.4\ \text{T}$,

$\omega = 10\pi\ \text{rad/s}$,

$R = 20\ \text{cm} = 0.2\ \text{m}$,

we substitute these values into the formula:

First calculate $R^2$:

$ R^2 = (0.2\ \text{m})^2 = 0.04\ \text{m}^2 $

Substitute into the formula:

$ V = \frac{1}{2} \times 0.4 \times (10\pi) \times 0.04 $

Multiply step-by-step:

$0.4 \times 10\pi = 4\pi$,

Then, $4\pi \times 0.04 = 0.16\pi$,

Finally, multiply by $\frac{1}{2}$:

$ V = \frac{1}{2} \times 0.16\pi = 0.08\pi $

Substitute $\pi \approx 3.14$:

$ V \approx 0.08 \times 3.14 = 0.2512\ \text{V} $

Thus, the potential difference developed between the axis and the rim is approximately $0.2512\ \text{V}$, which corresponds to Option C.

Let's analyze each statement:

$\textbf{A. The self-inductance of the coil depends on its geometry.}$

This is true because the self-inductance of a coil is given by formulas that include its geometrical parameters (number of turns, cross-sectional area, length, etc.). For example, for a solenoid,

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

where $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length.

$\textbf{B. Self-inductance does not depend on the permeability of the medium.}$

This is false. The permeability of the medium (represented by $\mu = \mu_0 \mu_r$) directly influences the inductance, as seen in the formula above. So, any change in the medium’s permeability will affect the inductance.

$\textbf{C. Self-induced e.m.f. opposes any change in the current in a circuit.}$

This is true, and it is a direct consequence of Lenz's law. The induced electromotive force (e.m.f.) always acts in a direction such that it opposes the change in current that produced it.

$\textbf{D. Self-inductance is the electromagnetic analogue of mass in mechanics.}$

This is true. Just as mass resists changes in velocity (inertia), inductance resists changes in current, which is why it is often compared to the inertial mass in mechanical systems.

$\textbf{E. Work needs to be done against self-induced e.m.f. in establishing the current.}$

This is true because, when you try to change the current, you must do work to overcome the opposing self-induced e.m.f., storing energy in the magnetic field of the inductor.

Based on the above reasoning, the true statements are A, C, D, and E.

Thus, the correct answer is:

Option C

A, C, D, E only

Motional emf : $\varepsilon=\mathrm{B} \ell \mathrm{v}=\mathrm{constant}$

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.

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.

$ \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.

$\phi=\left(5 t^2+4 t+2\right) \times 10^{-3}$ weber

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

$ \begin{aligned} & =\frac{d}{d t}\left(5 t^2+4 t+2\right) \times 10^{-3} \\ & =(10 t+4) \times 10^{-3} \mathrm{volt} \\ \text { At } t & =6 \mathrm{~s}, e=(10 \times 6+4) \times 10^{-3} \\ & =64 \times 10^{-3} \mathrm{volt} \\ & =6.4 \times 10^{-2} \mathrm{volt} \end{aligned} $

∴ Induced current

$ \begin{aligned} I & =\frac{e}{R}=\frac{6.4 \times 10^{-2}}{16} \\ & =0.4 \times 10^{-2} \mathrm{~A}=4 \times 10^{-3} \mathrm{~A}=4 \mathrm{~mA} \end{aligned} $

The small energy losses in transformers due to eddy currents can be reduced by laminating the core. The thin laminating core effectively increases the resistance to the flow of eddy currents within the core.

We know that the angle $\theta$ between the area of the coil and the direction of the magnetic field is $90^\circ$ when the coil is parallel to the field.

The formula for magnetic flux is $\phi = NBA\cos\theta$, where $N$ is the number of turns, $B$ is the magnetic field, $A$ is the area, and $\theta$ is the angle.

Substitute $\theta = 90^\circ$ into the formula. We get:

$\phi = N B A \cos 90^\circ = 0$

This is because $\cos 90^\circ = 0$.

$ \text { } \begin{aligned} |e| & =L \frac{d I}{d t} \\ 2.8 \times 10^{-3} & =L \frac{(8-3)}{0.2} \\ \Rightarrow \quad L & =\frac{2.8}{25} \times 10^{-3}=0.112 \times 10^{-3} \mathrm{H} \\ & =112 \times 10^{-6} \mathrm{H}=112 \mu \mathrm{H} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \text { Induced emf, } e=-N \frac{\Delta \phi}{\Delta t} \\ & =-\frac{900\left(\phi_2-\phi\right)}{\Delta t} \\ & =\frac{-900\left(B A \cos 180^{\circ}-B A \cos 0^{\circ}\right)}{\Delta t} \\ & =-900\left(\frac{-2 B A}{\Delta t}\right)=900 \times \frac{2 B A}{\Delta t} \\ & =\frac{900 \times 2 \times 3.5 \times 10^{-5} \times 3 \times 10^{-2}}{0.5} \\ & =3.78 \times 10^{-3} \mathrm{~V} \\ & \therefore \quad F=\frac{e}{R}=\frac{3.78 \times 10^{-3}}{1.8} \\ & =2.1 \times 10^{-3} \mathrm{~A}=2.1 \mathrm{~mA} \end{aligned} \end{aligned} $

Induced emf

$ \begin{aligned} E & =-\frac{N \Delta \phi}{\Delta t}=\frac{-N\left(\phi_2-\phi_1\right)}{\Delta t} \\ & =\frac{-N\left(B A \cos \theta_2-B A \cos \theta_1\right)}{\Delta t} \\ & =-250 \frac{\left(B A \cos 90^{\circ}-B A \cos 60^{\circ}\right)}{100 \times 10^{-3}} \\ & =\frac{-250\left(-\frac{B A}{2}\right)}{100 \times 10^{-3}} \\ & =2500 \times \frac{B A}{2}=1250 \times B A \\ & =1250 \times 2 \times 120 \times 10^{-4}=30 \mathrm{volt} \end{aligned} $

$\therefore$ Induced current

$ I=\frac{E}{R}=\frac{30}{8}=3.75 \mathrm{~A} $

$\phi=2 t^2+3 t-2$

$ \begin{aligned} & \therefore e=\frac{d \phi}{d t}=\frac{d}{d t}\left(2 t^2+3 t-2\right)=4 t+3 \\ & \text { at } t=3 \mathrm{~s}, e=4 \times 3+3 \\ & \quad e=15 \mathrm{volt} \end{aligned} $

Induced power, $P=\frac{e^2}{R}=\frac{15^2}{7.5}=30 \mathrm{~W}$

$ \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} $

Time taken to cross the field region

$=\frac{50}{2}=25 \mathrm{~s}$

At $10 \mathrm{~s}$ the loop is inside field and flux is not changing.

$\therefore \quad \varepsilon_{\text {induced }}=0$

To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf), which is given by Faraday's law of electromagnetic induction as it applies to self-induction:

$ \text{emf} = - L \frac{\Delta I}{\Delta t} $

Where:

• $ \text{emf} $ = induced voltage in volts (V)
• $ L $ = self-inductance of the coil in henries (H)
• $ \Delta I $ = change in current in amperes (A)
• $ \Delta t $ = time interval in seconds (s) over which the current change occurs

Here, the problem gives us the following data:

• $ \text{emf} = 0.1 \, \text{V} $
• $ \Delta I = 2 \, \text{A} - (-2 \, \text{A}) = 4 \, \text{A} $
• $ \Delta t = 0.2 \, \text{s} $

Substituting the given values into the formula, we get:

$ 0.1 = -L \frac{4}{0.2} $

Solving for $ L $, the equation becomes:

$ 0.1 = -L \cdot 20 $

Therefore:

$ L = - \frac{0.1}{20} $

Calculating the value of $ L $:

$ L = -0.005 \, \text{H} $

Or, expressing $ L $ in millihenries (mH):

$ L = -5 \, \text{mH} $

However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):

$ L = 5 \, \text{mH} $

Thus, the self-inductance of the coil is 5 mH, which corresponds to Option D.