Electromagnetic Induction

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} $

0 to L

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

L to 2 L

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

To find the energy stored in a coil with an inductance of 80 mH carrying a current of 2.5 A, we use the formula for the energy stored in an inductive coil:

$ U = \frac{1}{2} L i^2 $

Given:

Inductance, $ L = 80 \, \text{mH} = 80 \times 10^{-3} \, \text{H} $

Current, $ i = 2.5 \, \text{A} $

We substitute these values into the formula:

$ \begin{align*} U & = \frac{1}{2} \times 80 \times 10^{-3} \times (2.5)^2 \\ & = \frac{1}{2} \times 80 \times 10^{-3} \times 6.25 \\ & = 40 \times 10^{-3} \times 6.25 \\ & = 25 \times 10^{-2} \, \text{J} \\ U & = 0.25 \, \text{J} \end{align*} $

Therefore, the energy stored in the coil is $ 0.25 \, \text{J} $.

The mutual inductance between two coils is given as 8 mH, which can be converted to henries as:

$ M = 8 \, \text{mH} = 8 \times 10^{-3} \, \text{H} $

The current in one of the coils changes over time according to the equation:

$ I = 12 \sin(100t) $

where $ I $ is in amperes and $ t $ is in seconds.

To find the maximum emf induced in the second coil, we use Faraday's law of mutual induction, which states:

$ e = M \frac{dI}{dt} $

Calculating the derivative of the current with respect to time, we have:

$ \frac{dI}{dt} = \frac{d}{dt}[12 \sin(100t)] = 12 \times 100 \cos(100t) $

Substituting this into the formula for emf, we get:

$ \begin{align*} e &= M \cdot \frac{dI}{dt} \\\\ &= 8 \times 10^{-3} \times 12 \times 100 \cos(100t) \\\\ &= 9.6 \cos(100t) \end{align*} $

The expression $ e = 9.6 \cos(100t) $ is in the form of the standard equation for emf:

$ e = e_0 \cos(\omega t) $

From this, we can identify the maximum emf, $ e_0 $, as:

$ e_0 = 9.6 \, \text{V} $

Given,

Area of coil, $ A=200 \mathrm{~cm}^{2} $

Number of turn, $ N=50 $

Angular speed, $ \omega=40 \mathrm{rad} / \mathrm{s} $

Magnetic field, $ B=2 \times 10^{-2} \mathrm{~T} $

Maximum emf induced is given as $ \mathrm{emf}=N A B \omega \sin \omega t $

For maximum value $ \sin \omega t=1 $

$ \begin{aligned} (\mathrm{emf})_{\max } & =N A B \omega \\\\ & =50 \times 200 \times 10^{-4} \times 2 \times 10^{-2} \times 40 \\\\ & =0.8 \mathrm{~V} \end{aligned} $

Given, $l=1.66 \mathrm{~m}$

$ \begin{aligned} & \text { Speed }=90 \mathrm{kmph}=25 \mathrm{~m} / \mathrm{s} \\\\ & B_v=0.2 \times 10^{-4} \mathrm{~T} \\\\ & \text { Induced emf }=\text { Blv } \\\\ &=0.2 \times 10^{-4} \times 1.66 \times 25 \\\\ &=8.3 \times 10^{-4} \mathrm{~V} \\\\ &=0.83 \mathrm{mV} \end{aligned} $

Here, coil 1 has a radius of $r_1$ and coil 2 has a radius of $r_2$.

Given, $r_1 \ll r_2$

Let current $I_2$ flow in the coil 2 and magnetic field at its centre

$ B=\frac{\mu_0 I_2}{2 r_2} $

Since, $r_1 \ll r_2$

Flux through the coil, $\phi=\frac{\mu_0 I_2}{2 r_2} \times \pi \pi_1^2$

By definition of mutual inductance,

$ \phi=M_{12} I_2=\frac{\mu_0 I_2}{2 r_2} \pi r_1^2 \Rightarrow M_{12}=\frac{\mu_0}{2 r_2} \pi r_1^2 $

Hence, mutual inductance of the arrangement is $\frac{\mu_0}{2 r_2} \pi r_1^2$.

Given:

Initial current, $ I_1 = 14 \, \text{A} $

Final current, $ I_2 = 4 \, \text{A} $

Change in current, $ \Delta I = I_2 - I_1 = 4 \, \text{A} - 14 \, \text{A} = -10 \, \text{A} $

Time duration, $ \Delta t = 0.2 \, \text{ms} = 0.2 \times 10^{-3} \, \text{s} $

Average induced emf, $ \varepsilon = 150 \, \text{V} $

To find the self-inductance $ L $ of the circuit, we use the formula for induced emf in terms of self-inductance:

$ \varepsilon = L \left(\frac{dI}{dt}\right) $

Rearrange to solve for $ L $:

$ L = \frac{\varepsilon}{\left(\frac{dI}{dt}\right)} $

The rate of change of current $ \left(\frac{dI}{dt}\right) $ can be calculated as:

$ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{-10 \, \text{A}}{0.2 \times 10^{-3} \, \text{s}} $

Plug the values into the formula:

$ L = \frac{150 \, \text{V}}{\frac{-10 \, \text{A}}{0.2 \times 10^{-3} \, \text{s}}} $

Calculate:

$ L = \frac{150 \times 0.2 \times 10^{-3}}{-10} $

$ L = 3 \, \text{mH} $

Thus, the self-inductance of the circuit is $ 3 \, \text{mH} $.

$ \begin{aligned} &\text { Given, } l=0.1 \mathrm{~m}\\ &\begin{aligned} & \text { Resistance of wire }=1 \Omega \\ & \qquad \begin{array}{l} B=2 \mathrm{~Wb} / \mathrm{m}^2 \\ i=1 \mathrm{~mA}=10^{-3} \mathrm{~A} \end{array} \end{aligned} \end{aligned} $

Calculate equivalent resistance for Wheatstone bridge,

$ \begin{aligned} & R_1=3+3=6 \Omega \\ & R_2=3+3=6 \Omega \\ & R_{\mathrm{eq}}=\frac{R_1 \times R_2}{R_1+R_2}=\frac{6 \times 6}{6+6}=\frac{36}{12}=3 \Omega \end{aligned} $

So, total resistance in the circuit is,

$ R_{\text {total }}=3+1=4 \Omega $

The induced emf in the loop,

$ e=B v l $

So, induced current, $i=\frac{e}{R}=\frac{B v l}{R} \Rightarrow v=\frac{i R}{B l}$

$ \begin{aligned} & v=\frac{10^{-3} \times 4}{2 \times 0.1} \\ & v=2 \times 10^{-2} \mathrm{~m} / \mathrm{s}=2 \mathrm{~cm} / \mathrm{s} \end{aligned} $

Given that a coil with inductance $L$ is divided into six equal parts, these parts are connected in parallel. To find the resultant inductance of this parallel combination, we use the formula for inductors in parallel:

$ \frac{1}{L_{\text{parallel}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + \frac{1}{L_4} + \frac{1}{L_5} + \frac{1}{L_6} $

Since each part has equal inductance, we can express each as:

$ L_1 = L_2 = L_3 = L_4 = L_5 = L_6 = \frac{L}{6} $

Substituting into the parallel inductance formula gives:

$ \frac{1}{L_{\text{parallel}}} = 6 \times \frac{1}{L / 6} $

Simplifying this, we find:

$ L_{\text{parallel}} = \frac{L}{36} $

Therefore, the resultant inductance of the combination is $\frac{L}{36}$.

Given the following parameters:

Number of turns, $ N = 120 $

Inductance, $ L = 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H} $

Current, $ I = 30 \, \text{mA} = 30 \times 10^{-3} \, \text{A} $

The magnetic flux linked with the coil can be calculated using the formula:

$ \phi = \frac{L \cdot I}{N} $

Substituting the given values into the formula:

$ \phi = \frac{40 \times 10^{-3} \times 30 \times 10^{-3}}{120} $

Simplifying this:

$ \phi = 10 \times 10^{-6} \, \text{Wb} $

Thus, the magnetic flux linked with the coil is $ 10 \times 10^{-6} \, \text{Wb} $.

Given:

Induced current in circuit $Y$, $i_Y = 60 \times 10^{-4} \, \mathrm{A}$

Resistance of circuit $Y$, $R_Y = 4 \, \Omega$

Mutual inductance, $m = 3 \, \mathrm{mH} = 3 \times 10^{-3} \, \mathrm{H}$

Time, $t = 0.02 \, \mathrm{s}$

Formula:

The induced EMF ($\varepsilon_Y$) in circuit $Y$ is given by:

$ \left|\varepsilon_Y\right| = m \frac{d i_X}{d t} $

The current in circuit $Y$ due to the induced EMF can be given by:

$ i_Y R_Y = m \frac{d i_X}{d t} $

Calculation:

Substitute the known values into the equation:

$ \begin{aligned} d i_X &= \frac{60 \times 10^{-4} \times 4 \times 0.02}{3 \times 10^{-3}} \\ &= \frac{60 \times 10^{-4} \times 8 \times 10^{-2}}{3 \times 10^{-3}} \\ &= 160 \times 10^{-6} \times 10^{3} \\ &= 160 \times 10^{-3} \\ &= 0.16 \, \mathrm{A} \end{aligned} $

Thus, the change in current required in circuit $X$ is $0.16 \, \mathrm{A}$.

For solenoid $ A $:

Let $ n $ be the number of turns per unit length for solenoid $ A $, and $ l $ be its length.

The total number of turns $ N_A $ is given by:

$ N_A = n \cdot l $

For solenoid $ B $:

The number of turns per unit length for solenoid $ B $ is $ 3n $ (since the ratio is $ 1:3 $).

The length of solenoid $ B $ is $ 2l $.

The total number of turns $ N_B $ is:

$ N_B = (3n) \times (2l) = 6nl $

Self-Inductance Formulas:

The self-inductance $ L $ of a solenoid is given by the formula:

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

where $ \mu_0 $ is the permeability of free space, $ N $ is the number of turns, $ A $ is the cross-sectional area, and $ l $ is the length.

For solenoid $ A $:

$ L_A = \frac{\mu_0 (N_A)^2 A}{l} = \frac{\mu_0 (n \cdot l)^2 A}{l} = \mu_0 n^2 l A $

For solenoid $ B $:

$ L_B = \frac{\mu_0 (N_B)^2 A}{l} = \frac{\mu_0 (6n \cdot l)^2 A}{2l} = \frac{\mu_0 36n^2 l^2 A}{2l} = 18 \mu_0 n^2 l A $

Ratio of Self-Inductances:

Now, the ratio of the self-inductances $ \frac{L_A}{L_B} $ is:

$ \frac{L_A}{L_B} = \frac{\mu_0 n^2 l A}{18 \mu_0 n^2 l A} = \frac{1}{18} $

Thus, the ratio of the self-inductances of solenoids $ A $ and $ B $ is $ 1:18 $.

To find the magnitude of the induced electric field at a distance $ r $ from the $ Z $-axis in the presence of a time-dependent magnetic field, we begin with the magnetic flux:

Magnetic Flux

Given that the magnetic field $ \mathbf{B} $ is expressed as $ B = B_0 r t \hat{\mathbf{k}} $, we can calculate the magnetic flux $ \phi $ through a circular region of radius $ r $:

$ \phi = \int \mathbf{B} \cdot d\mathbf{A} = \int B \cdot 2 \pi r \, dr $

Substituting the expression for $ B $:

$ \phi = 2 \pi B_0 t \int r^2 \, dr = 2 \pi B_0 t \frac{r^3}{3} $

Induced Electric Field

According to Faraday’s law of electromagnetic induction, the induced electromotive force (EMF) $\oint \mathbf{E} \cdot d\mathbf{l}$ is related to the rate of change of magnetic flux:

$ \oint \mathbf{E} \cdot d\mathbf{l} = \frac{d\phi}{dt} $

Substituting the expression for $\phi$:

$ E(2 \pi r) = \frac{d}{dt}\left(2 \pi B_0 t \frac{r^3}{3}\right) $

Differentiating with respect to time $ t $:

$ E(2 \pi r) = \frac{2 \pi B_0 r^3}{3} $

Solving for $ E $:

$ E = \frac{2 \pi B_0 r^3}{3 \times 2 \pi r} = \frac{B_0 r^2}{3} $

Therefore, the magnitude of the induced electric field at a distance $ r $ from the $ Z $-axis is given by:

$ E = \frac{B_0 r^2}{3} $

The magnetic energy stored in an inductor is given by the formula:

$ U = \frac{1}{2} L I^2 $

Where:

$ U $ is the energy stored,

$ L $ is the inductance,

$ I $ is the current through the inductor.

In this case, the current increases from $ I_1 = 2 \, \mathrm{A} $ to $ I_2 = 3 \, \mathrm{A} $.

Calculating the initial energy stored $ U_1 $ when the current is $ I_1 $:

$ U_1 = \frac{1}{2} L I_1^2 = \frac{1}{2} \times L \times 4 = 2L $

Calculating the energy stored $ U_2 $ when the current is $ I_2 $:

$ U_2 = \frac{1}{2} L I_2^2 = \frac{1}{2} \times L \times 9 = 4.5L $

The increase in energy stored, $\Delta U$, is:

$ \Delta U = U_2 - U_1 = 4.5L - 2L = 2.5L $

To find the percentage increase in energy stored:

$ \text{Percentage Increase} = \frac{\Delta U}{U_1} \times 100\% = \frac{2.5L}{2L} \times 100\% = 125\% $

The changing magnetic field induces an electric field, which accelerates the electron.

To determine the acceleration of the electron, we use the formula:

$ a = \frac{eE}{m} $

where $ F = qE $.

To calculate the electric field $ E $, we apply Faraday's law of electromagnetic induction:

$ \varepsilon = -\frac{d\Phi}{dt} = -B \cdot A $

Here, $ B $ is the magnetic field, and $ A $ is the cross-sectional area. Therefore, the electromotive force (emf) is:

$ \varepsilon = -A \cdot \frac{dB}{dt} = -\pi r^2 \cdot \frac{dB}{dt} $

The electric field $ E $ is then given by:

$ E = -\frac{\varepsilon}{\text{distance}} = -\frac{\varepsilon}{d} $

$ E = \frac{\pi r^2}{d} \cdot \frac{dB}{dt} $

Finally, the acceleration $ a $ is:

$ a = \frac{e}{m} \cdot \frac{\pi r^2}{d} \cdot \frac{dB}{dt} $

Substituting the given values:

$ \begin{aligned} a &= \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \times \frac{22}{7} \times \frac{(0.5)^2}{0.3} \times 50 \times 10^{-6} \\ &= 23 \times 10^6 \, \mathrm{ms}^{-2} \end{aligned} $

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$