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A copper rod of mass $\mathrm{m}$ 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 $\mathrm{B}$ normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is:
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Verified Answer
The correct answer is:
$\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$
$\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$
From Faraday's law of electomagnetic induction,
$$
\begin{aligned}
e &=\frac{d \phi}{d t}=\frac{d(B A)}{l t}=\frac{d(B l l)}{d t} \\
&=\frac{B d l \times l}{d t}=B V l
\end{aligned}
$$
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
$$
m g \sin \theta=\frac{B^2 l V}{R} \Rightarrow V=\frac{m g R \sin \theta}{B^2 l^2}
$$
$$
\begin{aligned}
e &=\frac{d \phi}{d t}=\frac{d(B A)}{l t}=\frac{d(B l l)}{d t} \\
&=\frac{B d l \times l}{d t}=B V l
\end{aligned}
$$
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
$$
m g \sin \theta=\frac{B^2 l V}{R} \Rightarrow V=\frac{m g R \sin \theta}{B^2 l^2}
$$
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