A straight conducting rod of length 30cm 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 current of 3 $μ$ A is induced through the rod. The angle of dip will be :

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$\tan^{- 1} (\frac{3}{4})$

$\tan^{- 1} (\frac{1}{\sqrt{3}})$

$\tan^{- 1} (\sqrt{3})$

$\tan^{- 1} (\frac{1}{3})$

answer is D.

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Detailed Solution

Only veritical lines of earth’s magnetism will cut when the rod is allowed to slide over the rails. Let the vertical component of earth’s magnetic field is B_v .

Induced
current $i = \frac{B_{v} / v - 0}{R} = \frac{B_{v} / v}{R}$
$B_{V} = 10^{- 5} T Given B_{H} = 0.3 \times 10^{- 4} T$
The angle of dip $θ$ is given by $θ = \tan^{- 1} (\frac{B_{v}}{B_{H}})$