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