For a coil having \(L=2~\text{mH},\) the current flow through it is \(I=t^2e^{-t}.\) The time at which emf becomes zero is:
1. \(2\) s
2. \(1\) s
3. \(4\) s
4. \(3\) s

A square of side \(L\) meters lies in the \(XY\text-\)plane in a region where the magnetic field is given by \(\vec{B}=B_{0}\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )\text{T}\) where \(B_{0}\) is constant. The magnitude of flux passing through the square will be:
1. \(2 B_{0} L^{2}~\text{Wb}\)
2. \(3 B_{0} L^{2}~\text{Wb}\)
3. \(4 B_{0} L^{2}~\text{Wb}\)
4. \(\sqrt{29} B_{0} L^{2}~\text{Wb}\)$$

Two identical conductors \(P\) and \(Q\) are placed on two frictionless (conducting) rails \(R\) and \(S\) in a uniform magnetic field directed into the plane. If \(P\) is moved in the direction as shown in the figure with a constant speed, then rod \(Q\):

With the decrease of current in the primary coil from \(2\) A to zero in \(0.01\) s, the emf generated in the secondary coil is \(1000~\text{V}\). The mutual inductance of the two coils is:
1. \(1.25\) H
2. \(2.50\) H
3. \(5.00\) H
4. \(10.00\) H

A long solenoid has self-inductance \(L\). If its length is doubled keeping total number of turns constant, then its new self-inductance will be:
1. \(\frac{L}{2}\)
2. \(2L\)
3. \(L\)
4. \(\frac{L}{4}\)

An inductor is connected to a direct voltage source through a switch. Then: