A copper rod of mass m slides under gravity on two smooth parallel rails $l$ distance apart and set at an angle $ heta$ to the horizontal. At the bottom, the rails are joined by a resistance R, figure. There is a uniform magnetic field B perpendicular to the plane of the rails. The terminal velocity of the rod is [image:7_image11.jpeg]

Question

A copper rod of mass m slides under gravity on two smooth parallel rails $l$ distance apart and set at an angle $	heta$ to the horizontal. At the bottom, the rails are joined by a resistance R, figure. There is a uniform magnetic field B perpendicular to the plane of the rails. The terminal velocity of the rod is [image:7_image11.jpeg]

Dr. Abubacker · Accepted Answer

(c) Terminal velocity of the rod is attained when magnetic force on the rod $(Bil)$ balances the component of weight of the rod$(mg\sin{	heta)},\ figure$.
[image:7_image52.emf]
$ie\ ,\ \ Bil = mg\sin	heta$
$B\left( \frac{e}{R} ight)l = mg\sin	heta$
$\frac{Bl}{R}(e) = \ mg\sin	heta$
$\frac{Bl}{R}\left( Blv_{r} ight) = mg\sin	heta$
$r_{T} = \ \frac{mg\ R\sin	heta}{B^{2}l^{2}}$

Explanation

Reviewed by Dr. Abubacker

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