A conducting wire of length l is placed on a rough horizontal surface, where a uniform horizontal magnetic field B perpendicular to the length of the wire exists. Least values of the forces required to move the rod when a current 'i' is established in the rod are observed to be $F_{1}$ and $F_{2} (< F_{1})$ for the two possible directions of the current through the rod, respectively. Find the weight of the rod and the coefficient of friction between the rod and the surface

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Solution

Changing the direction of current in the wire, we can change the normal reaction on the wire by the surface.
In one case, magnetic force on the wire will be in upward direction while in the other case, it will be in the downward direction.
Hence, normal reaction, $N = m g \pm B i l$ .
$f_{(friction limiting)} = μ (m g \pm B i l)$
As $F_{1} > F_{2}$
$F_{1} = μ (m g + B i l)$ (i)
$F_{2} = μ (m g - B i l)$ (ii)
From equations (i) and (ii),
$\frac{F_{1}}{F_{2}} = \frac{m g + B i l}{m g - B i l}$
$m g = B i l [\frac{F_{1} + F_{2}}{F_{1} - F_{2}}]$ (iii)
From equations (i) and (iii), we get $μ = \frac{F_{1} - F_{2}}{2 B i l}$

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