JEE PYQ: Electromagnetic Induction Question 14
Question 14 - 2019 (08 Apr Shift 1)
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 strip is 50 grams, its resistance $10;\Omega$ and air drag negligible, N will be close to:
(1) 1000
(2) 50000
(3) 5000
(4) 10000
Show Answer
Answer: (3)
Solution
This is a case of damped oscillation. The damping coefficient comes from the magnetic braking: $F = -kx - \frac{B^2l^2}{R}v$. The amplitude decays as $A = A_0 e^{-\frac{B^2l^2}{2mR}t}$. For amplitude to decrease by factor $e$: $t = \frac{2mR}{B^2l^2} = \frac{2 \times 50 \times 10^{-3} \times 10}{0.01 \times 0.01} = 10000$ s. Time period $T = 2\pi\sqrt{\frac{m}{k}} = 2$ s. Number of oscillations $N = \frac{10000}{2} = 5000$.
BETA
