Q: Two concentric coplanar circular loops have diameters 20 cm and 2 m and resistance of unit length of the wire = 10^{–4} Ω/m. A time -dependent voltage V = (4 + 2.5 t) volts is applied to the larger as shown. Find the current in the smaller loop .

Solution : r_{1} = 1.0 m , r_{2} = 10^{–1} m

Resistance of outer loop= 2 π × 10^{–4} Ω

Resistance of inner loop = 0.2 π × 10^{–4} Ω

Current in outer loop $\displaystyle = \frac{V}{R} = \frac{(4 + 2.5 t)}{2\pi \times 10^{-4}} A $

$\displaystyle I = \frac{(4 + 2.5 t)}{2\pi \times 10^{-4}} A $

Magnetic field produced at the common centre is

$\displaystyle B = \frac{\mu_0 I}{2 r_1} $

$\displaystyle B = \frac{4\pi \times 10^{-7}}{2} \times \frac{(4 + 2.5 t)}{2\pi \times 10^{-4}} $

B = 2 × 10^{-3} (2 + 1.25 t) T

Hence, flux linked with the inner loop,

φ = BA = 2 × 10^{–3} (2 + 1.25 t) × π(0.1)^{2}

= 2 π × 10^{–5} (2 + 1.25 t) Wb

Hence, the e.m.f. induced in smaller loop

$\displaystyle e = – \frac{d\phi}{dt}$

e = – 2 π × 10^{–5} (1.25 )

e = – 2.5 π × 10^{–5} Volt

The negative sign indicates that the induced e.m.f.(or current) is opposite to applied e.m.f. (or current) Hence, the current induced in the inner (smaller) loop is

$\displaystyle i = \frac{|e|}{R} = \frac{2.5 \pi \times 10^{-5}}{0.2 \pi \times 10^{-4}}$

i = 1.25 A