Question

A straight conducting rod of length $30$ cm and having a resistance of $0.2$ ohm is allowed to slide over two parallel thick metallic rails with uniform velocity of $0.2$ m/s as shown in the figure. The rails are situated in a horizontal plane. If the horizontal component of earth's magnetic field is $0.3\times {10}^{-4}$T and a steady state current of $3\mu A$ is induced through the rod. The angle of dip will be :

• A. $ta{n}^{-1}\left(\frac{3}{4}\right)$
• B. $ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. $ta{n}^{-1}\left(\sqrt{3}\right)$
• D. $ta{n}^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is D $ta{n}^{-1}\left(\frac{1}{3}\right)$

Only vertical lines of earth's magnetic field will cut when the rod is allowed to slide over the rails.
Let the vertical compents of the earth's magnetic field is $B_v$.
Induced current $i=\frac{B_{v}lv}{R}$

$3\times {10}^{-6}=\frac{B_v\times 0.3\times 0.2}{0.2}$
$B_v={10}^{-5}T$
given $B_H=0.3\times {10}^{-4}T$
The angle of dip $\theta$ is given by $=ta{n}^{-1}\left(\frac{B_v}{B_H}\right)=ta{n}^{-1}\left(\frac{1}{3}\right)$