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PhysicsInteger+4Question 45 of 69

IIT-JEE 2011 · Paper 1, Physics question 45

A long circular tube of length 10 m and radius 0.3 m carries a current II along its curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as I=I0cos⁡(300t)I=I_0\cos(300t) where I0I_0 is constant. If the magnetic moment of the loop is Nμ0I0sin⁡(300t)N\mu_0I_0\sin(300t), then 'NN' is

A vertical cylindrical tube with its axis dashed; a small coaxial wire loop inside it; an arrow on the curved surface shows the current I flowing round the tube (azimuthally).
Answer and worked solution

Official answer: 6

Field in the tube. The current on the curved surface acts like a solenoid sheet with current per length I10\frac I{10}: B=μ0I10B=\dfrac{\mu_0I}{10}.

Induced current. Flux Φ=π(0.1)2μ0I10=10−3πμ0I\Phi=\pi(0.1)^2\dfrac{\mu_0I}{10}=10^{-3}\pi\mu_0I, so
ε=−dΦdt=0.3πμ0I0sin⁡(300t),i=ε0.005=60πμ0I0sin⁡(300t).\varepsilon=-\frac{d\Phi}{dt}=0.3\pi\mu_0I_0\sin(300t),\qquad i=\frac\varepsilon{0.005}=60\pi\mu_0I_0\sin(300t).

Moment. iA=60π×0.01π μ0I0sin⁡(300t)=0.6π2μ0I0sin⁡(300t)≈6μ0I0sin⁡(300t)iA=60\pi\times0.01\pi\,\mu_0I_0\sin(300t)=0.6\pi^2\mu_0I_0\sin(300t)\approx6\mu_0I_0\sin(300t).

Answer: 6

Topics:Faraday's and Lenz's lawsAmpere's circuital law
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