Correct Option is (A)

From energy conservation

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

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

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

Correct Option is (A)

$\begin{array}{rl}& \mathrm{EMF}=\frac{\mathrm{d}\varphi }{\mathrm{dt}}=\frac{\mathrm{BA}-0}{\mathrm{t}}\\ \\ & \mathrm{A}=\pi {\mathrm{r}}^2=\pi \left(\frac{{0.1}^2}{\pi }\right)=0.01\\ \\ & \mathrm{B}=0.5\\ \\ & \mathrm{EMF}=\frac{(0.5)(0.01)}{0.5}=0.01\mathrm{V}=10\mathrm{mV}\end{array}$

Correct answer is 60

The induced emf in the circuit is given by Faraday's law of electromagnetic induction, which is $\mathcal{E}=-d\varphi /dt$, where $\varphi$ is the magnetic flux through the circuit.

The magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\varphi =NBA$, where $N$ is the number of turns in the loop, $B$ is the magnetic field through the core, and $A$ is the cross-sectional area of the core.

As the magnetic field changes from $1.5\mathrm{T}$ in one direction to $-1.5\mathrm{T}$ in the opposite direction, the change in magnetic flux is $\mathrm{\Delta }\varphi =2NBA$.

The induced emf drives a current $I$ through the resistor in the circuit, and the current and the resistance are related by Ohm's law, which is $I=\mathcal{E}/R$. Substituting the expression for $\mathcal{E}$ into this equation, we get $I=-d\varphi /dtR$.

The charge $Q$ that flows through the circuit during the change in magnetic field is given by $Q=\int Idt$. Substituting the expression for $I$ into this equation and integrating with respect to time, we get $Q=-\mathrm{\Delta }\varphi /R$, where $\mathrm{\Delta }\varphi$ is the change in magnetic flux and $R$ is the resistance of the circuit.

Substituting the given values into this expression, we get:

$Q=-\frac{2NBA}{R}=-\frac{2(100)(1.5)(24\times {10}^{-4})}{12}=-0.06\mathrm{C}=-60\mathrm{mC}$

Therefore, the charge flowing through a point in the circuit during the change of magnetic field is $60\mathrm{mC}$, which is the same as the provided answer.

Correct answer is 50

The problem involves a conducting circular loop placed in a uniform magnetic field with its plane perpendicular to the field. The radius of the loop is expanding at a constant rate, and we are asked to find the magnitude of the induced emf in the loop at an instant when the radius of the loop is $2\mathrm{cm}$.

The magnetic flux through a circular loop of radius $r$ and area $A=\pi r^2$ placed in a uniform magnetic field $B$ perpendicular to the plane of the loop is given by:

${\mathrm{\Phi }}_B=BA=B\pi r^2$

The induced emf in the loop is given by Faraday's law of electromagnetic induction:

$\mathcal{E}=-\frac{d{\mathrm{\Phi }}_B}{dt}$

In this case, the radius of the loop is expanding at a constant rate of ${10}^{-3}\mathrm{m}/\mathrm{s}$, which means that the rate of change of the area of the loop is:

$\frac{dA}{dt}=\frac{d}{dt}(\pi r^2)=2\pi r\frac{dr}{dt}=2\pi (0.02\mathrm{m})({10}^{-3}\mathrm{m}/\mathrm{s})=4\times {10}^{-5}{\mathrm{m}}^2/\mathrm{s}$

The magnetic flux through the loop is changing at this rate, and the induced emf in the loop is given by:

$\mathcal{E}=\left|\frac{d{\mathrm{\Phi }}_B}{dt}\right|=\left|\frac{dB}{dt}\frac{dA}{dt}\right|=\left|B\frac{dA}{dt}\right|=|0.4\mathrm{T}\times 4\times {10}^{-5}{\mathrm{m}}^2/\mathrm{s}|=16\pi \mu \mathrm{V}$

Therefore, the magnitude of the induced emf in the loop at an instant when the radius of the loop is $2\mathrm{cm}$ is $50.24$ $\simeq$ 50 $\mu \mathrm{V}$.

Correct answer is 8

$\begin{array}{rl}& \mathrm{m}=\tan \theta =\frac{10}{5}=2\\ \\ & \mathrm{B}=\mathrm{mt}\\ \\ & \mathrm{B}=2\mathrm{t}\end{array}$

$\begin{array}{rl}& \epsilon =\left|\frac{\mathrm{d}\varphi }{\mathrm{dt}}\right|=\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=\frac{\mathrm{AdB}}{\mathrm{dt}}\\ \\ & \epsilon =\frac{4\mathrm{d}(2\mathrm{t})}{\mathrm{dt}}=4\times 2=8\mathrm{mV}\end{array}$