The magnetic field in the cylindrical region shown in figure increases at a constant rate of 20.0 mT/s. Each side of the square loop abcd and defa has a length of 1.00 cm and a resistance of 4.00 Ω. Find the current (magnitude and sense) in the wire ad if (a) the switch S_{1}_{ }is closed but S_{2} is open, (b) S_{1} is open but S_{2} is closed, (c) both S_{1} and S_{2} are open and (d) both S_{1} and S_{2} are closed.

#### Solution

(a) When switch S_{1} is closed and switch S_{2} is open:-

Rate of change of the magnetic field = 20 mT/s = \[2 \times {10}^{- 4} T/s\]

Net resistance of the coil adef, R = 4 × 4 = 16 Ω

Area of the coil adef = (10^{−2} )^{−2} = 10^{−4} m^{2}

The emf induced is given by

\[e = \frac{d\phi}{dt} = A . \frac{dB}{dt}\]

= 10^{−4} × 2 × 10^{−2}

= 2 × 10^{−6} V

The current through the wire ad is given by

\[i= \frac{e}{R} = \frac{2 \times {10}^{- 6}}{16}\]

= 1.25 × 10^{−7} A along ad

(b) When switch S_{2} is closed and switch S_{1} is open:-

Net resistance of the coil abcd, R = 16 Ω

The induced emf is given by

\[e = A \times \frac{dB}{dt} = 2 \times {10}^{- 6} V\]

The current through wire ad is given by

\[i = \frac{20 \times {10}^{- 6}}{16} = 1 . 25 \times {10}^{- 7} A \] along da

(c) When both S_{1} and S_{2} are open, no current is passed, as the circuit is open. Thus, i = 0.

(d) When both S_{1} and S_{2} are closed, the circuit forms a balanced a Wheatstone bridge and no current flows along ad. Thus, i = 0.