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Q Type your question. negligible resistance (Fig 3.126). The rails are connected to each other at the bottom by a resistanceless rail paralle to the wire so that the wire and the rails form a closed rectangular conducting loop. The plane of the rails makes an angle \( \theta \) with the horizontal and a uniform vertical magnetic field of induction B exist throughout the region. Find the steady-state velocity of the wire.

1. \( m g=\sin \theta \)
2. \( \frac{m g}{R} \frac{\sin ^{2} \theta}{B^{2} l^{2} \cos ^{2} \theta} \)
3. \( \frac{m g R \sin \theta}{B^{2} l^{2} \cos ^{2} \theta} \)
4. \( \operatorname{mgr} \frac{\sin ^{2} \theta}{B^{2} / 2 \cos \theta} \)

Correct answer: \( \frac{m g R \sin \theta}{B^{2} l^{2} \cos ^{2} \theta} \)

Solution

At steady speed, the net force along the incline is zero, so the component of gravity down the plane is balanced by the magnetic force opposing motion. The moving rod induces an emf, which drives a current through resistance R; that current in the magnetic field produces a force proportional to B, l, and the rod’s speed, leading to the stated expression.

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