JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Electromagnetic Induction

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A thin strip 10 cm long is on a ∪ 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

10Ω

and air drag negligible, N will be close to:

(A)50000
(B)10000
(C)1000
(D)5000✓ Correct

Answer: (D)

Solution

The moving strip is damped by eddy-current drag, so its amplitude decays as

A = A_{0} e^{- t / γ}

with time constant

γ = \frac{2 m}{b}

, where the electromagnetic damping coefficient is

b = \frac{B ^{2} ℓ ^{2}}{R}

. Here

b = \frac{( 0.1 ) ^{2} ( 0.1 ) ^{2}}{10} = 1 0^{- 4}

kg/s, so

γ = \frac{2 \times 0.05}{1 0 ^{- 4}} = 1 0^{4}

s is the time for the amplitude to fall by a factor

e

. The period is

T_{0} = 2 π \frac{m}{k} = 2 π \frac{0.05}{0.5} = \frac{2 π}{10}

s, so the number of swings in that time is

N = \frac{γ}{T _{0}} = \frac{1 0 ^{4} 10}{2 π} \approx 5000

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Electromagnetic Induction

Solution

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