A copper rod of length L and mass m is sliding down a smooth inclined plane of inclination $θ$ with a constant
speed v. A current I is flowing in perpendicular to the plane of diagram inwards. A vertically upward magnetic
field $\to B$ exist in this region.The magnitude of the required magnetic field $\to B$ is

$\frac{m g}{I L} s i n θ$

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$\frac{m g}{I L} s i n θ$

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$\frac{m g}{I L} t a n θ$

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$\frac{m g}{I L s i n θ}$

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Solution

The correct option is C
$\frac{m g}{I L} t a n θ$

BIL $c o s θ = m g s i n θ$ or $B = \frac{m g}{I L} \frac{s i n θ}{c o s θ} = \frac{m g}{I L} t a n θ$

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A copper rod of length $L$ and mass $m$ carrying current $I$ flowing inwards and perpendicular to the plane(as shown in the figure) , is placed on a fixed smooth inclined plane of inclination $θ$ . If a vertically upward magnetic field $\to B$ exists in this region, find the magnitude of the magnetic field $\to B$ for which the copper rod slides down the inclined plane with a constant speed.