InterviewSolution
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InterviewSolution
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A metallic rod of linear density is 0.25 kg m^(-1) is lylong horizontally on a smooth inclined plane which makes an angle of 45^(@) with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of strength 0.25 T is acting on it in the vertical direction. Calculates the electric current flowing in the rod to keep it stationary . |
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Answer» Solution :The linear density of the rod I mass per UNTI length of the rod is 0.25 kg `m^(-1)` `rArr (m)/(l) = 0.25 kg m^(-1)`. Let l be the current flowing in the matallic rod. The direction of electron current is into the paper . The direction of magnetic force IB/is given by Fleming.s LEFT hand rule For equilibrum , mg sin `45^(@) - IB/ cos 45^(@)` `rArr 1 = (l)/(B) (m)/(l) G tan 45^(@) = (0.25 kg m^(-1))/(0.25 l )= 1 xx 9.8 m s^(-2)` `rArr`I = 9.8 A so, we NEED to supply current of `9 xx` A to keep the METALLIC rod stationary. 
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