Question

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 S₁ is closed but S₂ is open, (b) S₁ is open but S₂ is closed, (c) both S₁ and S₂ are open and (d) both S₁ and S₂ are closed.

Answer 1

(a) When switch S₁ is closed and switch S₂ is open:
Rate of change of the magnetic field = 20 mT/s = $2\times {10}^{-4}\mathrm{T}/\mathrm{s}$
Net resistance of the coil adef, R = 4 × 4 = 16 Ω
Area of the coil adef = (10⁻² )⁻² = 10⁻⁴ m²
The emf induced is given by
$e=\frac{d\mathrm{\varphi }}{dt}=\mathrm{A}.\frac{d\mathrm{B}}{dt}$
= 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 S₂ is closed and switch S₁ is open:
Net resistance of the coil abcd, R = 16 Ω
The induced emf is given by
$e=\mathrm{A}\times \frac{d\mathrm{B}}{dt}=2\times {10}^{-6}\mathrm{V}$
The current through wire ad is given by
$i=\frac{20\times {10}^{-6}}{16}=1.25\times {10}^{-7}\mathrm{A}$ along da
(c) When both S₁ and S₂ are open, no current is passed, as the circuit is open. Thus, i = 0.
(d) When both S₁ and S₂ are closed, the circuit forms a balanced a Wheatstone bridge and no current flows along ad. Thus, i = 0.