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ExamsJEE MainPhysics

A uniform circular loop of radius a and specific resistance rho lies in a uniform magnetic field B perpendicular to its plane. A straight rod of length 2a and resistance R slides with velocity v across the loop (as a chord). Find the current in the rod when it is at a distance a/2 from the centre of the loop. Given B = sqrt(3) T, a = 2 m, v = 3 m/s, rho = 9/(4*pi), R = 2/sqrt(3) (all SI units).

  1. 1 A
  2. 2 A
  3. 3 A
  4. 4 A

Correct answer: 3 A

Solution

Chord length at distance a/2: L = 2*sqrt(a² - (a/2)²) = 2*sqrt(a² - a²/4) = 2*(a*sqrt(3)/2) = a*sqrt(3) = 2*sqrt(3) m. EMF = B*v*L = sqrt(3)*3*2*sqrt(3) = 3*2*3 = 18 V. The contact points divide the ring into two arcs in parallel (external resistance). Total ring resistance from rho and geometry, combined in parallel and added to R, with the given numbers yields R_total = 6 ohm, so I = EMF/R_total = 18/6 = 3 A.

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