Figure shows two long metal rails placed horizontally and parallel to each other at a separation $l$ . A uniform magnetic field B exists in the vertically downward direction. A wire of mass m can slide on the rails. The rails are connected to a constant current source which drives a current i in the circuit. The friction coefficient between the rails and the wire is µ . What should be the minimum value of µ which can prevent the wire from sliding on the rails?

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$\frac{ilB}{mg}$

$\frac{2 ilB}{mg}$

$\frac{3 ilB}{mg}$

$\frac{4 ilB}{mg}$

answer is A.

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Detailed Solution

The force on the wire due to the magnetic field is $\vec{F} = i \vec{l} \times \vec{B}$
Or, $F = ilB$
It acts towards right in the given figure. If the wire does not slide on the rails, the force of friction by the rails should be equal to F. If $μ_{0}$ be the minimum coefficient of friction which can prevent sliding, this force is also equal to $μ_{0}$ mg. Thus,
$μ_{0} mg = ilB Or, μ_{0} = \frac{ilB}{mg}$