Correct answer is 44

In this problem, a square loop is inside a long solenoid, and there's a varying current flowing through the solenoid. Because the current is changing, it induces a changing magnetic field inside the solenoid.

According to Faraday's law of electromagnetic induction, a changing magnetic field will induce an electromotive force (emf) in a loop placed in that field. In this case, the loop is the square loop inside the solenoid.

The formula used here is based on Faraday's law, which states that the induced emf in a loop is equal to the rate of change of magnetic flux through the loop. This is given by:

$\text{emf}=-\frac{d\mathrm{\Phi }}{dt}$

where $\mathrm{\Phi }$ is the magnetic flux.

The magnetic field inside a solenoid is given by $B={\mu }_{0}nI$, where ${\mu }_0$ is the permeability of free space, $n$ is the number of turns per unit length in the solenoid, and $I$ is the current through the solenoid.

The magnetic flux through the square loop is then given by $\mathrm{\Phi }=B\cdot A={\mu }_{0}nIA$, where $A$ is the area of the loop.

When the current is sinusoidal, i.e., $I(t)=I_0\sin (\omega t)$, its derivative with respect to time is $dI/dt=I_0\omega \cos (\omega t)$, where $\omega$ is the angular frequency.

Hence, the rate of change of flux becomes:

$\frac{d\mathrm{\Phi }}{dt}={\mu }_{0}nA\frac{dI}{dt}={\mu }_{0}nA{I}_0\omega \cos (\omega t)$

The emf, which is equal to the negative of the rate of change of flux, will have a maximum value (the amplitude) when $\cos (\omega t)=1$, giving:

$\text{Emf amplitude}={\mu }_{0}nA{I}_0\omega$

$=4\pi \times {10}^{-7}\text{T m/A}\times \left(\frac{50}{{10}^{-2}}\right)\text{turns/m}\times (2\times {10}^{-2}\text{m}{)}^2\times 2.5\text{A}\times 700\text{rad/s}$

which simplifies to:

$\text{Emf amplitude}=44\times {10}^{-4}\text{V}$

So, the value of $x$ in the question is $44$

Correct answer is 10

$EMF=\frac{d}{dt}\left(B\pi r^2\right)$

$=2\mathrm{B}\pi \mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}=2\times \pi \times 0.1\times 0.8\times 2\times {10}^{-2}$

$=2\pi \times 1.6=\mathbf{1}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{6}[$ rounding off $\mathbf{1}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{6}=\mathbf{1}\mathbf{0}]$

Correct Option is (C)

$\begin{array}{rl}\mathrm{P}& =\frac{e^2}{\mathrm{R}}\\ \\ e& =\text{ induced emf }=\frac{\mathrm{N}d{\varphi }_B}{dt}=\mathrm{NA}\frac{d\mathrm{B}}{dt}\end{array}$

$\text{ and }\mathrm{R}=\frac{el}{\pi r^2}\text{ where }r\text{ is the radius of the wire }$

$\begin{array}{ll}& \mathrm{P}=\frac{{\left(\mathrm{NA}\frac{d\mathrm{B}}{dt}\right)}^2}{\frac{el}{\pi r^2}}\\ \\ \Rightarrow & \mathrm{P}\propto {\mathrm{N}}^2{r}^2\\ \\ \Rightarrow & {\mathrm{P}}^{\prime }\propto {\left(\frac{\mathrm{N}}{2}\right)}^2(2r{)}^2\\ \\ \Rightarrow & {\mathrm{P}}^{\prime }=\mathrm{P}\end{array}$

Correct Option is (A)

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

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

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

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

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

Correct answer is 12

$B_{\mathrm{\perp }}=3t^2$

$\frac{d{B}_{\mathrm{\perp }}}{dt}=6t-12$ at $t=2$

$\frac{d{\varphi }_1}{dt}=12\times \pi (1{)}^2=12\pi$