A copper rod of mass m slides under gravity on two smooth paralle… - Vidyadip

MCQ

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

A
$\frac{{mgR\cos \,\theta }}{{{B^2}{l^2}}}$
✓
$\frac{{mgR\sin \,\theta }}{{{B^2}{l^2}}}$
C
$\frac{{mgR\tan \,\theta }}{{{B^2}{l^2}}}$
D
$\frac{{mgR\cot \,\theta }}{{{B^2}{l^2}}}$

✓

Answer

Correct option: B.

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

b
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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