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 figure. The terminal speed of the copper rod is

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Answer & Solution

Correct Answer

(B) \(\frac{{mgR\sin \,\theta }}{{{B^2}{l^2}}}\)

Detailed explanation

From Faraday's law of electomagnetic induction,

\(e=\frac{d \phi}{d t}=\frac{d(B A)}{l t}=\frac{d(B l l)}{d t}\)

\( = \frac{{Bdl \times l}}{{dt}} = BVl\)

Also, \(F=i l B=\left(\frac{B V}{R}\right)\left(l^{2} B\right)=\frac{B^{2} l^{2} V}{R}\)

At equilibrium

\(m g \sin \theta=\frac{B^{2} I V}{R} \Rightarrow V=\frac{m g R \sin \theta}{B^{2} l^{2}}\)

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