Electromagnetic Induction

JEE Main level, single correct, medium

95 of 153

Question 95 of 153

A thin strip 10cm10cm long is on a UU shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5Nm−10.5N{m}^{-1} (see figure). The assembly is kept in a uniform magnetic field of 0.1T.0.1T. 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 ee is NN . If the mass of the strip is 5050 grams, its resistance 10Ω10\Omega and air drag negligible, NN will be close to:

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Answer: (B) 50005000

Equation of motion for slider:
−md2xdt2=kx+ilB-m\dfrac{{d}^{2}x}{d{t}^{2}}=kx+ilB
md2xdt2+kx+(BlvR)lB=0⇒md2xdt2+B2l2R.dxdt+kx=0m\dfrac{{d}^{2}x}{d{t}^{2}}+kx+\left(\dfrac{Blv}{R}\right)lB=0\Rightarrow m\dfrac{{d}^{2}x}{d{t}^{2}}+\dfrac{{B}^{2}{l}^{2}}{R}.\dfrac{dx}{dt}+kx=0
Comparing with
md2xdt2+b.dxdt+kx=0m\dfrac{{d}^{2}x}{d{t}^{2}}+b.\dfrac{dx}{dt}+kx=0
m=m;b=B2l2R;k=km=m;b=\dfrac{{B}^{2}{l}^{2}}{R};k=k
We know:
A=A0e−b2m×t=A0e−B2l22mR×tA={A}_{0}{e}^{\dfrac{-b}{2m}\times t}={A}_{0}{e}^{-\dfrac{{B}^{2}{l}^{2}}{2mR}\times t}
For amplitude to be
A0/e,{A}_{0}/e,
A0e−B2l22mR×t=Ae⇒B2l22mR×t=1{A}_{0}{e}^{-\dfrac{{B}^{2}{l}^{2}}{2mR}\times t}=\dfrac{A}{e}\Rightarrow \dfrac{{B}^{2}{l}^{2}}{2mR}\times t=1
t=2mRB2l2=10,000sect=\dfrac{2mR}{{B}^{2}{l}^{2}}=10,000sec
Time period =
2πmk=2s2\pi \sqrt{\dfrac{m}{k}}=2s
Number of oscillations =
10,0002=5,000\dfrac{10,000}{2}=5,000

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