$A$ 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 $\frac{dB}{dt} \text{ (T/s)}$. $A$ proton of charge $e$ and mass $m$ is placed at point $P$ on the periphery. Its acceleration is

The alternating e.m.f. induced in the secondary coil of a transformer is mainly due to

$A$ 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.

$A$ square loop of side $2 \text{ cm}$ is placed in a time-varying magnetic field with magnitude $B = 0.4 \sin(300t) \text{ T}$. The normal to the plane of the loop makes an angle of $60^{\circ}$ with the field. The maximum induced emf produced in the loop is . . . . . . $\text{mV}$.

$A$ conducting ring of radius $r$ is placed perpendicularly inside a time-varying magnetic field given by $B = B_0 + \alpha t$ as shown in the figure. $B_0$ and $\alpha$ are positive constants. Find the $emf$ produced in the ring.

The circular wire in the figure below encircles a solenoid in which the magnetic flux is increasing at a constant rate out of the plane of the page. The clockwise emf around the circular loop is $\varepsilon_{0}$. By definition,a voltmeter measures the voltage difference between two points given by $V_{b}-V_{a}=-\int_{a}^{b} E \cdot ds$. We assume that $a$ and $b$ are infinitesimally close to each other. The values of $V_{b}-V_{a}$ along path $1$ and $V_{a}-V_{b}$ along path $2$,respectively,are