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)`.

A. `0.01A`
B. `0.1A`
C. `0.02A`
D. `0.2A`

Answer 1

Correct Answer - A
Force exerted by air on the rod
`=rho(L//3xx2R)upsilon^(2)=rhoLRupsilon^(2)`
Balancing torque about `O`,
`NI(piR^(2))=(rhoLRupsilon^(2))xx(3L)/(4)`
`implies300pi/BR=(3rhoupsilon^(2)L^(2))/(4)`
`implies1=(1)/(400)xx(rhoL^(2)upsilon^(2))/(piBR)=0.01A`.