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A rectangular wire frame ABCDA has resistance per unit length equal to R/(2*l). The longer sides (AB and CD) have length 2*l and the shorter sides (BC and DA) have length l. The frame moves with constant velocity V0 along two fixed, smooth, horizontal, resistanceless conducting rails in a uniform magnetic field B perpendicular to the plane of the rails. Only one long side of the frame cuts the field lines (one side is the leading edge generating EMF). Which of the following statements about the moving frame are correct?

1. External force required to move the frame is 2*B²*l²*V0 / (5*R).
2. External force required to move the frame is 4*B²*l²*V0 / (5*R).
3. Potential difference between points B and C is zero.
4. Potential difference between points B and C is 2*B*V0*l.

Correct answer: External force required to move the frame is 4*B²*l²*V0 / (5*R).

Solution

The induced EMF = B*l*V0 from the leading side. The frame's circuit gives two parallel paths. Computing the equivalent resistance and current, the external force equals 4*B²*l²*V0/(5R).

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