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A metallic rod of mass per unit length 0...

A metallic rod of mass per unit length 0.5 kg `m^(-1)` is lying horizontally on a smooth inclined plane which makes an anlge 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

A

14.76A

B

5.98 A

C

7.14 A

D

11.32 A

Text Solution

Verified by Experts

The correct Answer is:
D
Metallic rod is at rest on inclined plane. Let mass of the rod is m and mass per unit length is M, then
`M =(m)/(l) = 0.5 kg//m`
The net force on the rod is zero.

Current flows the horizontal direction (perpendicular to plane of paper) and magnetic field is vertical then choosing appropriate direction of current magnetic force (IIB) should be horizontal as shown in figure. For equilibrium of rod we can write the following equation.
`mg sin30^(@)= Fcos30^(@)`
`IlB= mg tan30^(@)`
`I= (mg)/(lB)tan30^(@)= (Mg)/(B)tan30^(@)`
`= (0.5 xx 9.8 )/(0.25 xx sqrt(3))= 11.32A`
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