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ExamsJEE AdvancedPhysics

A long circular tube of length 10 m and radius 0.3 m carries a current I along its curved surface (acting as a long solenoid-like sheet). A wire loop of resistance 0.005 ohm and radius 0.1 m is placed coaxially inside the tube. The current varies as I = I0*cos(300t). If the induced magnetic moment of the loop is N*mu0*I0*sin(300t), find N.

  1. 6
  2. 3
  3. 12
  4. 1

Correct answer: 6

Solution

Field inside the tube: B = mu0*(I/L) (single current sheet of length L carrying total I, treated as n = 1/L turns per metre). Flux through loop: phi = B*pi*r² = mu0*(I/L)*pi*r². emf = -dphi/dt = mu0*(pi*r²/L)*I0*300*sin(300t). Induced current i = emf/Res. Magnetic moment m = i*pi*r² = (mu0*pi²*r⁴*300/(L*Res))*I0*sin(300t). Plug r = 0.1, L = 10, Res = 0.005: pi²*r⁴ = pi²*1e-4; factor = mu0*I0*sin(300t)*(pi²*1e-4*300/(10*0.005)) = mu0*I0*sin(300t)*(pi²*1e-4*300/0.05) = mu0*I0*sin(300t)*(pi²*0.6) approx mu0*I0*sin(300t)*5.92, so N approx 6.

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