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Earth's Magnetism

A copper rod ...

Question

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

A

\frac{m g R tan θ}{B^{2} l^{2}}

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B

\frac{m g R cot θ}{B^{2} l^{2}}

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C

\frac{m g R sin θ}{B^{2} l^{2}}

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D

\frac{m g R cos θ}{B^{2} l^{2}}

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Solution

The correct option is C $\frac{m g R sin θ}{B^{2} l^{2}}$
$\begin{matrix} L e t V i s t e r min a l s p e e d E m f = B l V i = \frac{B l V}{R} F_{B} = B (\frac{B l V}{R}) l = \frac{B^{2} l^{2} V}{R} m g sin θ = \frac{B^{2} l^{2} V}{R} V = \frac{m g R sin θ}{B^{2} l^{2}} \end{matrix}$
Hence,
option $(C)$ is correct answer.

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