Q: 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° with the horizontal. The rod is not allowed to slide down by flowing a current through it when 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) 14.76 A

(b) 11.32 A

(c) 5.98 A

(d) 7.14 A

Ans: (b)

Sol: For stationary rod ,

$ \displaystyle i l B cos\theta = m g sin\theta $

$ \displaystyle i = \frac{m g sin\theta}{l B cos\theta} $

$ \displaystyle i = \frac{m g tan\theta}{l B } $

$ \displaystyle i = \frac{0.5 \times 10 \times tan30}{1 \times 0.25 } $

i = 11.32 A