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 : r1 = 1.0 m , r2 = 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