$A$ circular loop of radius $20 \text{ cm}$ and resistance $2 \text{ } \Omega$ is placed in a time-varying magnetic field $\vec{B} = (2t^2 + 2t + 3) \text{ T}$. The plane of the loop is perpendicular to the magnetic field. The induced current in the loop at $t = 3 \text{ s}$ is $\frac{\alpha}{50} \text{ A}$. The value of $\alpha$ is . . . . . . .

$A$ point charge $Q$ is moving in a circular orbit of radius $R$ in the $x$-$y$ plane with an angular velocity $\omega$. This can be considered as equivalent to a loop carrying a steady current $I = \frac{Q\omega}{2\pi}$. $A$ uniform magnetic field along the positive $z$-axis is now switched on,which increases at a constant rate from $0$ to $B$ in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that,for an orbiting charge,the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $\gamma$.
$1.$ The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is:
$(A)$ $\frac{BR}{4}$ $(B)$ $\frac{BR}{2}$ $(C)$ $BR$ $(D)$ $2BR$
$2.$ The change in the magnetic dipole moment associated with the orbit,at the end of the time interval of the magnetic field change,is:
$(A)$ $-\gamma BQR^2$ $(B)$ $-\gamma \frac{BQR^2}{2}$ $(C)$ $\gamma \frac{BQR^2}{2}$ $(D)$ $\gamma BQR^2$
Give the answer for question $1$ and $2$.

$A$ magnetic field given by $B(t) = (0.2t - 0.05t^2) \text{ T}$ is directed perpendicular to the plane of a circular coil containing $25$ turns of radius $1.8 \text{ cm}$ and whose total resistance is $5 \Omega$. The power dissipation at $3 \text{ s}$ is nearly: (in $\text{ } \mu\text{W}$)

In the branch $AB$ of a circuit,as shown in the figure,a current $I = (t + 2) \ A$ is flowing,where $t$ is the time in seconds. At $t = 0$,the value of $(V_A - V_B)$ will be: (in $V$)

$A$ uniform magnetic field of induction $B$ is confined to a cylindrical region of radius $R$. The magnetic field is increasing at a constant rate of $\frac{dB}{dt} \ (T/s)$. An electron is placed at the point $P$ on the periphery of the field. The acceleration experienced by the electron is:

$A$ uniform magnetic field $B$ exists in a cylindrical region of radius $R = 10 \, cm$ as shown in the figure. $A$ uniform wire of length $L = 80 \, cm$ and resistance $R_{wire} = 4.0 \, \Omega$ is bent into a square frame of side length $a = 20 \, cm$ and is placed with one side along a diameter of the cylindrical region. If the magnetic field increases at a constant rate of $\frac{dB}{dt} = 0.010 \, T/s$,find the current induced in the frame.