Correct option is (B)

We can apply Faraday's law of electromagnetic induction to solve this problem. Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.

The cube is moving through a magnetic field, so it's behaving like a conductor moving through a magnetic field. The induced emf or voltage can be calculated by using the formula:

emf = B $\times$ v $\times$ d

Where:

B is the magnetic field strength,

v is the velocity of the conductor, and

d is the length of the conductor perpendicular to the direction of motion and magnetic field.

In this case, the cube is moving in the y-direction and the magnetic field is in the z-direction. So, the faces perpendicular to the x-axis are involved. The length of the conductor (d) perpendicular to the motion and magnetic field is the edge length of the cube, which is 2 cm or 0.02 m.

So, plugging the given values into the formula:

emf = 0.1 T $\times$ 6 m/s $\times$ 0.02 m = 0.012 V = 12 mV

So, the potential difference between the two faces of the cube perpendicular to the x-axis is 12 mV.

Therefore, Option B is correct.

Correct option is (A)

At    t  =  0 s

B(0) = 0.4 sin (0) = 0

and at t  =  10 ms

B(10) = 0.4 sin (50$\pi$$\times$10$\times$10^-3)

= 0.4 sin $\left(\frac{\pi }{2}\right)$

= 0.4

As q = $\frac{\mathrm{\Delta }\varphi }{R}$

= $\frac{A\left[B\left(10\right)-B\left(0\right)\right]}{10}$

= $\frac{3.5\times {10}^{-3}\left[0.4-0\right]}{10}$

= 0.14 mC

Correct option is (C)

Flux in the coil, $\varphi$ = nBA sin($\omega$t)

When n = no. of turns

$$ A = Area of coil

$$$\omega$ = angular speed

Induced emf,

$\left|e\right|=\frac{d\varphi }{dt}$

= nBA$\omega$ cos$\omega$t

$\therefore$ e_max = nBA$\omega$

Correct option is (A)

We know that electric flux $\varphi =\overrightarrow{B}.\overrightarrow{A}$

$\Rightarrow \varphi =BA\cos \omega t$

Now, $B=\frac{{\mu }_0}{2}\frac{I}{R}$ is magnetic field due to circular coil of radius R and $A=\pi r^2$ is area of circular coil of radius r. Therefore,

$\varphi =\frac{{\mu }_0}{2}\frac{I}{R}\pi r^2\cos \omega t$

Now induced emf $\epsilon =\frac{-d\varphi }{dt}=\frac{-d}{dt}\left(\frac{{\mu }_0}{2}\frac{I}{R}\pi r^2\cos \omega t\right)$

$\Rightarrow \epsilon =\frac{{\mu }_0}{2}\frac{I}{R}\pi r^2\sin \omega t$

Correct option is (D)

According to Faraday's law of electromagnetic induction,

$\epsilon =\frac{d\varphi }{dt}$

Also, $\epsilon$ = iR

$\therefore$ $\frac{d\varphi }{dt}$ = iR

$\Rightarrow$ $\int d\varphi =R\int idt$

Magnitude of change in flux (d$\varphi$) = R × area under current vs time graph

$\Rightarrow$ d$\varphi$ = $100\times \frac{1}{2}\times \frac{1}{2}\times 10$ = 250 Wb