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).

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

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

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

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.

$\begin{aligned} & \frac{\phi_{A / B}}{I_B}=M \\ & \Rightarrow \quad M=\frac{\mu_0 I_B \times \pi a^2}{2 b \times I_B} \\ & =\frac{\mu_0 \pi a^2}{2 b} \end{aligned}$

To find the current in the secondary coil of the transformer, we first need to calculate the output power, taking into account the efficiency. The efficiency ($ \eta $) of the transformer is given by the ratio of the output power ($ P_{\text{out}} $) to the input power ($ P_{\text{in}} $) times 100%.

The given efficiency is $ \eta = 80\% $ or $ \eta = 0.8 $ in decimal form. The input power is also given as $ P_{\text{in}} = 4 \text{kW} $ or $ P_{\text{in}} = 4000 \text{W} $.

Let's calculate $ P_{\text{out}} $ using the efficiency formula:

$ P_{\text{out}} = \eta \times P_{\text{in}} = 0.8 \times 4000 \text{W} = 3200 \text{W} $

Now that we have $ P_{\text{out}} $, we can calculate the secondary current ($ I_{\text{secondary}} $) using the formula:

$ P_{\text{out}} = V_{\text{secondary}} \times I_{\text{secondary}} $

We are given $ V_{\text{secondary}} = 240 \text{V} $.

Isolating $ I_{\text{secondary}} $ gives us:

$ I_{\text{secondary}} = \frac{P_{\text{out}}}{V_{\text{secondary}}} $

Substituting the known values:

$ I_{\text{secondary}} = \frac{3200 \text{W}}{240 \text{V}} $

$ I_{\text{secondary}} = \frac{3200}{240} $

$ I_{\text{secondary}} = 13.33 \text{A} $

Therefore, the current in the secondary coil is $ 13.33 \text{A} $, which corresponds to Option B.

$\begin{aligned} \varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \\ \Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\ \mathrm{B}_{\mathrm{i}} & =5000 \mathrm{~T}, \\ \mathrm{~B}_{\mathrm{f}} & =3000 \mathrm{~T} \\ \mathrm{~d} & =0.02 \mathrm{~m} \\ \mathrm{r} & =0.01 \mathrm{~m} \\ \Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\ & =(2000) \pi(0.01)^2=0.2 \pi \\ \varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \Rightarrow 22=\mathrm{N}\left(\frac{0.2 \pi}{2}\right) \\ \mathrm{N} & =70 \end{aligned}$

Let's identify each law listed in List I and match it with the corresponding mathematical expression listed in List II.

$ \oint \vec{B} \cdot \vec{da} = 0 $

So, (A) matches with (II).

$ \oint \vec{E} \cdot \vec{dl} = -\frac{d}{dt} \int \vec{B} \cdot \vec{da} $

Hence, (B) matches with (III).

$ \oint \vec{B} \cdot \vec{dl} = \mu_0 I $

Consequently, (C) matches with (IV).

Lastly, Gauss's law of electrostatics states that the total electric flux out of a closed surface is proportional to the charge enclosed within the surface:

$ \oint \vec{E} \cdot \vec{da} = \frac{1}{\varepsilon_0} \int \rho dV $

Therefore, (D) matches with (I).

Based on these matches, the correct answer must link A-II, B-III, C-IV, and D-I:

The correct option is:

Option C

A-II, B-III, C-IV, D-I

Ampere-Maxwell law

$\rightarrow \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dl}}=\mu_0 \mathrm{i}_{\mathrm{c}}+\mu_0 \varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}$

Faraday law $\rightarrow \oint \vec{E} \cdot \overrightarrow{d l}=\frac{d \phi_{\mathrm{B}}}{d t}$

Gauss' law for electricity $\rightarrow \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\frac{\mathrm{Q}}{\varepsilon_0}$

Gauss ' law for magnetism $\rightarrow \oint \vec{B} \cdot \overrightarrow{d A}=0$

According to Faraday's Law of Electromagnetic Induction, the induced emf in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, it is expressed as:

$ \varepsilon = -\frac{d\Phi}{dt} $

Where $ \varepsilon $ is the induced emf, and $ \Phi $ is the magnetic flux.

To find the magnetic flux $ \Phi $ through the rectangular loop, we use the formula:

$ \Phi = B \cdot A \cdot \cos(\theta) $

Where:

• $ B $ is the magnetic field strength,
• $ A $ is the area of the loop, and
• $ \theta $ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.

The area $ A $ of the rectangular loop is:

$ A = \text{length} \times \text{width} = 2.5 \mathrm{~m} \times 2 \mathrm{~m} = 5 \mathrm{~m}^2 $

Given the angle $ \theta = 60^\circ $, we can calculate the initial magnetic flux $ \Phi_{initial} $:

$ \Phi_{initial} = B \cdot A \cdot \cos(60^\circ) $

$ \Phi_{initial} = 4 \mathrm{~T} \cdot 5 \mathrm{~m}^2 \cdot \cos(60^\circ) $

$ \Phi_{initial} = 4 \cdot 5 \cdot \frac{1}{2} = 10 \mathrm{~Wb} $

(Since $ \cos(60^\circ) = \frac{1}{2} $)

When the loop is removed from the magnetic field, the final magnetic flux $ \Phi_{final} $ is zero, because the loop is no longer within the magnetic field. Thus, the change in magnetic flux $ \Delta\Phi $ is:

$ \Delta\Phi = \Phi_{final} - \Phi_{initial} = 0 - 10 \mathrm{~Wb} = -10 \mathrm{~Wb} $

The loop is removed from the field in $ t = 10 $ seconds, so the rate of change of magnetic flux is:

$ \frac{d\Phi}{dt} = \frac{\Delta\Phi}{\Delta t} = \frac{-10 \mathrm{~Wb}}{10 \mathrm{~s}} = -1 \mathrm{~Wb/s} $

Now we can find the average induced emf $ \varepsilon $:

$ \varepsilon = -\frac{d\Phi}{dt} $

$ \varepsilon = -(-1 \mathrm{~Wb/s}) $

$ \varepsilon = +1 \mathrm{~V} $

Therefore, the average induced emf in the loop during this time is $ +1 \mathrm{~V} $. The correct answer is:

Option C

$ +1 \mathrm{~V} $

Assertion A is true because a bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with the same geometry and mass. This is due to the effect of the magnet's magnetic field on the metallic pipe.

Reason R is also true because when the magnetic bar moves through the metallic pipe, it induces a changing magnetic field in the pipe. This changing magnetic field, in turn, induces Eddy currents in the pipe. According to Lenz's law, these Eddy currents produce their own magnetic field, which opposes the motion of the magnetic bar, causing it to fall more slowly through the pipe.

Since both Assertion A and Reason R are true and R provides the correct explanation for A, the correct answer is Both A and R are true and R is the correct explanation of A.

Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.

This statement is true. The formula for current sensitivity ($I_s$) of a moving coil galvanometer is given by:

$I_s = \frac{NAB}{k}$

where:

• $N$ is the number of turns in the coil,
• $A$ is the area of the coil,
• $B$ is the magnetic field strength, and
• $k$ is the spring constant of the coil.

From this formula, you can see that the current sensitivity is directly proportional to the number of turns (N). If $N$ is doubled, then the current sensitivity will also double.

Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

This statement is false. The formula for voltage sensitivity ($V_s$) of a moving coil galvanometer is given by:

$V_s = I_s R = \frac{NAB}{k} R$

where:

• $R$ is the resistance of the coil.

From this formula, you can see that the voltage sensitivity is proportional to the number of turns ($N$) but also inversely proportional to the coil resistance ($R$). If you double the number of turns ($N$), you also double the length of the wire making up the coil, and thus, you double the resistance ($R$) of the coil. The doubling of $N$ is offset by the doubling of $R$, so the overall voltage sensitivity remains the same.

Therefore, increasing the current sensitivity by only increasing the number of turns in the coil will not increase the voltage sensitivity in the same ratio.

The emf induced in a rod moving through a magnetic field is given by Faraday's law of electromagnetic induction, specifically, in the form of motional emf, which states that:

$ \text{emf} = B \cdot L \cdot v $

where:

• (B) is the magnetic field strength,
• (L) is the length of the rod, and
• (v) is the velocity of the rod.

In this case, we are given the emf, (B), and (L), and we need to solve for (v). Rearranging the equation gives:

$ v = \frac{\text{emf}}{B \cdot L} $

Substituting the given values:

$ v = \frac{0.08 \, \text{V}}{0.4 \, \text{T} \times0.1 \, \text{m}} = 2 \, \text{m/s} $

The function of the non-magnetic metallic core in a galvanometer is to provide a path for the induced current generated by the moving coil in the magnetic field. This induced current opposes the motion of the coil and brings it to rest quickly due to the effect known as eddy current damping.

When the coil swings past its equilibrium position, a change in magnetic flux occurs. According to Faraday's law of electromagnetic induction, this change in magnetic flux generates an induced current, known as an eddy current. The eddy current creates its own magnetic field which opposes the original change in flux, causing the coil to quickly come to rest.

From energy conservation

Work done to pull the loop out = Energy is lost in the resistance

Emf in the loop $ = {{d\phi } \over {dt}} = {{B \times A} \over t} = {{40 \times 25 \times {{10}^{ - 4}}} \over {1s}} = 0.1\,V$

Energy lost $ = {{em{f^2}} \over R} = {{{{(0.1)}^2}} \over {10}} = {10^{ - 3}}\,J$

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

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

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

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

$ \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