A thin strip 10 cm long is on a U shaped wire of negligible… | JEE Main 2019 (8 Apr, Shift 1)

Q12JEE Main 2019 · 8 APR · SHIFT 1Difficult

Other: Induction & Displacement Current

A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 Nm–1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10 $\Omega$ and air drag negligible, N will be close to:

Show answer and solution

Answer: (c)

Induced current gives a damping force $F = \frac{B^2 l^2}{R} v = bv$ , with $b = \frac{(0.1)^2 (0.1)^2}{10} = 10^{-5}$ kg/s.
Amplitude $A = A_0 e^{-bt/2m}$ falls by $e$ when $t = \frac{2m}{b} = \frac{2 \times 0.05}{10^{-5}} = 10^4$ s.
Period $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.05}{0.5}} \approx 1.99$ s
$N = \frac{t}{T} = \frac{10^4}{1.99} \approx 5000$ , so the answer is (c).

Report a mistake

Other: Induction & Displacement Current question from JEE Main 2019 (8 Apr, Shift 1)

More questions on Other: Induction & Displacement Current