Question

Consider a variation of the previous problem (figure). Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistances but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity ω in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle θ with the vertical.

Answer 1

Magnitude of Force is $\frac {B^2a^2\omega}{2R} - mg sin\theta$

Explanation:

As we know that,

• F = $\frac {B^2a^2\omega}{2R}$ = ilB

and the loop along with rod have Negligible resistances

i = $\frac {Ba^2\omega^2}{2R}$

• Force on rod is given by -

$F_B = iBl = i = \frac {Ba^2 \omega^2}{2R}$

• Net Force = $F - \frac {Ba^2\omega^2}{2R} + mg sin\theta$

Component of mg along F = $mgsin\theta$

Net Force = $\frac {B^2a^2\omega}{2R} - mg sin\theta$