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AB and CD are two frictionless parallel rails a distance l apart, inclined at angle theta to the horizontal. A uniform magnetic field of magnitude B points vertically upward throughout the region. A conducting rod EF of mass m, carrying current i, rests across the rails. For EF to remain in equilibrium on the incline, which condition holds?

1. Current must flow from E to F
2. B*i*l = m*g*tan(theta)
3. B*i*l = m*g*sin(theta)
4. B*i*l = m*g

Correct answer: B*i*l = m*g*tan(theta)

Solution

Since B is vertical and the rod (current direction) is horizontal, the magnetic force F = B*i*l is horizontal. For equilibrium on the incline, resolve along the slope: the horizontal force's component up the incline, F*cos(theta), must balance the gravity component down the incline, m*g*sin(theta). So B*i*l*cos(theta) = m*g*sin(theta), giving B*i*l = m*g*tan(theta).

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