A fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self-inductance $L$ at $t=0.$The time at which the energy is stored equally between the electric and the magnetic fields is:

A metallic rod of length $l$ is tied to a string of length $2l$ and made to rotate with angular speed $\omega$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field $B$ in the region, the emf induced across the ends of the rod is:

In the circuit shown here, the point $C$is kept connected to point $A$ till the current flowing through the circuit becomes constant. Afterward, suddenly point $C$ is disconnected from point $A$ and connected to point $B$ at time $t=0.$Ratio of the voltage across resistance and the inductor at $t=\frac{L}{R}$ will be equal to:

An inductor $\left(L=0.03 \mathrm{H}\right)$ and a resistor $\left(R=0.15 \mathrm{k\Omega }\right)$ are connected in series to a battery of $15 \mathrm{V}$ emf in a circuit shown below. The key $K_1$ has been kept closed for a long time. Then at $t=0$, $K_1$ is opened and key $K_2$ is closed simultaneously. At $t=1 \mathrm{ms}$, the current in the circuit will be $\left(e^5\cong 150\right)$

In a coil resistance $100 \mathrm{\Omega },$ a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is:

A metallic coil of $N$ turns of radius $a$, resistance $R$, and inductance $L$ is held fixed with its axis along a spatial uniform magnetic field $B$ whose magnitude is given by $B_{0 }\sin \left(\omega t\right).$

(a) Write the emf equation for the current $i$ in the coil.

(b) Assuming that in the steady-state $i$ oscillates with the same frequency $\mathrm{\omega }$ as the magnetic field, obtain the expression for $i$.

(c) Obtain the force per unit length. Further, obtain its oscillatory part and the time-averaged compressional part.

(d) Calculate the time-averaged compressional force per unit length given that $B_0=1.00 \mathrm{tesla}, N=10, a=10.0 \mathrm{cm}, \mathrm{\omega }=1000.0 \mathrm{rad}-{\mathrm{s}}^{–1}, R =10.0 \mathrm{\Omega }, L= 100.0 \mathrm{mH}.$