Question

A straight conducting rod of length $30\text{cm}$ and having a resistance of $0.2\Omega$ is allowed to slide over two parallel thick metallic rails with uniform velocity of $0.2{\text{ms}}^{-1}$ as shown in the figure. The rails are stacked in a horizontal plane. If the horizontal component of earth's magnetic field is $0.3\times {10}^{-4}\text{T}$ and a steady current of $3\mu \text{A}$ is induced through the rod. The angle of dip will be

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

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{1}{3}\right)$

Only vertical component of earth's magnetism will pass through the rod when it is allowed to slide over the rails.

Let $B_v$ be the vertical component of earth`s magnetic field. $E=B_{v}lV$

Induced current $=i=\frac{E}{R}=\frac{B_{v}lV}{R}$

Substituting the data given in the question,
$B_v=\frac{Ri}{lV}=\frac{0.2\times 3\times {10}^{-6}}{30\times {10}^{-2}\times 0.2}={10}^{-5}\text{T}$

and $B_H=0.3\times {10}^{-4}\text{T}\text{or}3\times {10}^{-5}\text{T}$

We know , $\tan \theta =\frac{B_v}{B_H}=\frac{{10}^{-5}}{3\times {10}^{-5}}$

The dip angle is given by $\theta ={\tan }^{-1}\left(\frac{1}{3}\right)$

Hence, option (a) is the correct answer.