Question

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. 11.32 A
• B. 7.14 A
• C. 14.76 A
• D. 5.98 A

Answer 1

The Correct Option is A

For equilibrium,
mg $\sin 30^{\circ}=I / B \cos 30^{\circ}$
$I =\frac{ mg }{l B } \,\tan 30^{\circ}$
$=\frac{0.5 \times 9.8}{0.25 \times \sqrt{3}} $
$=11.32\, A$