Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $A=10 {\mathrm{cm}}^2$ and $\mathrm{length}=20 \mathrm{cm}.$ If one of the solenoids has $300 \mathrm{turns}$ and the other$400 \mathrm{turns},$their mutual inductance is $({\mathrm{\mu }}_0=4\mathrm{\pi }\times {10}^{–7} \mathrm{T} \mathrm{m} {\mathrm{A}}^{–1}):$

An inductor of inductance, $L=400 \mathrm{mH}$ and resistors of resistances $R_1=2 \mathrm{\Omega }$ and $R_2=2 \mathrm{\Omega }$ are connected to a battery of emf $12 \mathrm{V}$as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0.$The potential drop across $L$ as a function of time is:

A rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R \mathrm{\Omega }$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2 \mathrm{and} I$ are:

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: