Question

A non-conducting rod having circular cross section of radius R is suspended from a rigid support as shown in fig. A light and small coil of 300turns is wrapped tightly at the left end where uniform magnetic filed B exists in vertically downward direction. Air of density `rho` hits the half of the right part of the rod with velocity V as shown in the fig. What should be current in clockwise direction (as seen from O) in the coil so that rod remains horizontal? Give answer in mA. Given `2/(Lv)sqrt((piRB)/(rho))=1/(sqrt5)A^(-1//2)`.

Answer 1

Correct Answer - 2
Force exerted by air on the rod `=(rho 1/22R)v^2=rhoLRv^2`
Balancing torque about point `O, NI(piR^2)B=rho=LRv^2(3L)/4`
`implies 300pilBR =(3rhov^2L^2)/4`
`implies I=(rhoL^2v^2)/(400piBR)=1/100 ((Lv)/2 sqrt(rho/(piBR)))^2=0.002A=2mA`