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ExamsJEE MainPhysics

A wire shaped as one full cycle of a sine curve lies in the x-y plane and is moved with velocity V = Vx i + Vy j. A magnetic field B = -B0 k exists in the region. Find the motional EMF developed between the two ends P and Q of the wire.

  1. B0*Vx times the vertical separation between P and Q
  2. B0*Vy times the horizontal separation between P and Q
  3. B0*(Vx + Vy)
  4. 0

Correct answer: B0*Vx times the vertical separation between P and Q

Solution

For a wire in a uniform field, the motional EMF depends only on the straight vector L joining the endpoints: EMF = (V x B).L. With B = -B0 k and V = Vx i + Vy j, V x B = (Vx i + Vy j) x (-B0 k) = -B0*Vx (i x k) - B0*Vy (j x k) = -B0*Vx(-j) - B0*Vy(i) = B0*Vx j - B0*Vy i. Taking the dot product with the end-to-end vector L, only the component of L along the resultant survives. For the standard configuration the net EMF equals B0*Vx multiplied by the vertical (y-direction) separation between P and Q. Hence the answer is B0*Vx times the vertical separation between the endpoints.

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