Question 109: Electromagnetic Induction

Question

A copper rod of mass m slides under gravity on two smooth parallel rails, with separation
l
and set at an angle of $$	heta $$ 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 :

Accepted Answer

At terminal velocity, net force on rod = $$mg\sin 	heta $$
$$ \Rightarrow mg\sin 	heta  = iBl$$
Now, $$IBl = \left( {{{Bvl} \over R}} ight)Bl = {{{B^2}{l^2}v} \over R}$$
$$ \Rightarrow mg\sin 	heta  = {{{B^2}{l^2}v} \over R} \Rightarrow v = {{mgR\sin 	heta } \over {{B^2}{l^2}}}$$

Answer

$${{mg\,R\,	an \,	heta } \over {{B^2}\,{l^2}}}$$

Answer

$${{mg\,R\,\cot \,	heta } \over {{B^2}\,{l^2}}}$$

Answer

$${{mg\,R\,\sin \,	heta } \over {{B^2}\,{l^2}}}$$

Answer

$${{mg\,R\,\cos \,	heta } \over {{B^2}\,{l^2}}}$$

Electromagnetic Induction

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