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ExamsJEE AdvancedPhysics

A conducting rod of length l = 2*sqrt(5) m and mass m = 4 kg lies on a horizontal table. The coefficient of friction between the rod and the table is mu = 1/2. A current of 2 A flows through the rod. Find the minimum magnitude of the magnetic field (in tesla) required to make the rod just start translating along the x-axis. (g = 10 m/s²; neglect the radius of the rod.)

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

Correct answer: 2

Solution

The rod lies on a horizontal table. The magnetic force F = I*l x B. To move the rod along x-axis with minimum B, orient B in the horizontal plane perpendicular to the rod (maximise sin(theta)=1 if possible). But if B has a vertical component it changes the normal force. Optimal B is horizontal and perpendicular to the rod: F_horizontal = BIl, friction = mu*mg. So BIl = mu*mg => B = mu*mg/(I*l) = (1/2)*4*10/(2*2*sqrt(5)) = 20/(4*sqrt(5)) = 5/sqrt(5) = sqrt(5) ~ 2.24. But from the options, B = 2 T. Re-examine: if B is at angle phi to horizontal, vertical component B*sin(phi) changes normal force. F_horiz = I*l*B*cos(phi), N = mg - I*l*B*sin(phi), friction = mu*N. Condition: I*l*B*cos(phi) = mu*(mg - I*l*B*sin(phi)). B = mu*mg/(I*l*(cos(phi)+mu*sin(phi))). Minimise denominator over phi... actually maximise (cos(phi)+mu*sin(phi)) => tan(phi)=mu=1/2 => cos(phi)=2/sqrt(5), sin(phi)=1/sqrt(5). B_min = (1/2)*4*10 / (2*2*sqrt(5)*(2/sqrt(5)+1/2*1/sqrt(5))) = 20/(4*sqrt(5)*(2/sqrt(5)+1/(2*sqrt(5)))) = 20/(4*sqrt(5)*5/(2*sqrt(5))) = 20/(4*sqrt(5)*5/(2*sqrt(5))) = 20/(10) = 2 T.

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