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Q.

A non-conducting non-magnetic rod having circular cross section of radius R is suspended from a rigid support as shown in the figure. A light and small coil of 300 turns is wrapped tightly at the left end of the rod where uniform magnetic field B exists in vertically downward direction.Air of density ρ hits the half of the right part of the rod with velocity V as shown in the figure. What should be current in clockwise direction (as seen from O) in the coil so that rod remains horizontal? Give answer in m A. Given2LvπRBρ=15 A-1/2

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answer is 2.

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Detailed Solution

Force exerted by air on the rod =ρL22Rv2=ρLRv2  Balancing torque about point O,NIπR2B=ρLRv23L4 ⇒  300πIBR=3ρv2L24 ⇒  I=ρL2v2400πBR=1100Lv2ρπBR2=0.002 A=2 mA
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