A metallic rod of mass per unit length 0.5 kgm–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 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
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a
7.14 A
b
5.98 A
c
14.76 A
d
11.32 A
answer is D.
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Detailed Solution
Mass per unit length of a metallic rod is ml = 0.5 kg m−1.Let I be the current flowing. For equilibrium,mgsin30∘=IlBcos30∘ ⇒I=mglBtan30∘=0.5×9.80.25×3=11.32A