Moving Charges and Magnetism - Result Question 56
57. A metallic rod of mass per unit length $0.5 kg$ $m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25 T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
(a) $7.14 A$
(b) $5.98 A$
(c) $11.32 A$
(d) $14.76 A$
[2018]
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Answer:
Correct Answer: 57. (c)
Solution:
- (c) From figure, for equilibrium,
$ \begin{aligned} & \text{ (c) From figure, for equilibrium, } \\ & mg \sin 30^{\circ}=I / B \cos 30^{\circ} \\ & \Rightarrow \quad I=\frac{m g}{\ell B} \tan 30^{\circ} \\ & =\frac{0.5 \times 9.8}{0.25 \times \sqrt{3}}=11.32 A \xrightarrow{B^{\circ}} \end{aligned} $
