JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 7)
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A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of $$\theta $$ with the horizontal. At the bottom rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the igure. The terminal speed of the copper rod is :
$${{mg\,R\,\tan \,\theta } \over {{B^2}\,{l^2}}}$$
$${{mg\,R\,\cot \,\theta } \over {{B^2}\,{l^2}}}$$
$${{mg\,R\,\sin \,\theta } \over {{B^2}\,{l^2}}}$$
$${{mg\,R\,\cos \,\theta } \over {{B^2}\,{l^2}}}$$
Explanation
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At terminal velocity, net force on rod = $$mg\sin \theta $$
$$ \Rightarrow mg\sin \theta = iBl$$
Now, $$IBl = \left( {{{Bvl} \over R}} \right)Bl = {{{B^2}{l^2}v} \over R}$$
$$ \Rightarrow mg\sin \theta = {{{B^2}{l^2}v} \over R} \Rightarrow v = {{mgR\sin \theta } \over {{B^2}{l^2}}}$$
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