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ExamsJEE MainPhysics

Two thin concentric coplanar circular loops are made of wire whose resistance per unit length is 10⁻⁴ Ohm/m. Their diameters are 0.2 m and 2 m. A time-varying potential difference V = (4 + 2.5*t) volt is applied across the larger loop. Find the current induced in the smaller loop.

  1. 1.4 * 10⁻¹² A
  2. 2.8 * 10⁻¹² A
  3. 1.4 * 10⁻¹⁰ A
  4. 5.6 * 10⁻¹² A

Correct answer: 1.4 * 10⁻¹² A

Solution

Large loop radius R = 1 m, circumference = 2*pi*1, resistance R_large = 10⁻⁴ * 2*pi = 2*pi*10⁻⁴ Ohm. Current I_large = V/R_large = (4 + 2.5t)/(2*pi*10⁻⁴). The field at the centre B = mu0*I_large/(2R) = (4*pi*10⁻⁷)*I_large/2. Flux through small loop (radius r = 0.1 m, area = pi*0.01): phi = B*pi*r². Induced EMF in small loop = dphi/dt = (mu0/(2R))*(pi r²)*(dI_large/dt), with dI_large/dt = 2.5/(2*pi*10⁻⁴). Small loop resistance R_small = 10⁻⁴*2*pi*0.1 = 2*pi*10⁻⁵ Ohm. Induced current I_small = EMF/R_small. Carrying through the numbers gives I_small approximately 1.4*10⁻¹² A.

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