Question

Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. The centre O and a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly, so that the rod rotates clockwise at a uniform angular velocity ω. Find the force.
Figure

Answer 1

Force is equal to Product of Field, Current and Length

Explanation:

Let an element of rod of thickness, 'dr' at some distance 4 from the center of element.

• Now, the linear speed of the element at r from the center.  

de = Blv

de = $B \times dr \times \omega r$

e = $\int_{0}^{a} (B\omega r)dr$

= $\frac {1}{2} B\omega a^2$

• Because it is connected to resistance, the current in the circuit containing the rod, wire, and circular loop is given by $i = \frac {B\omega a^2}{2R}$

• The Direction of the current is from A to point O in the rod.

• The magnitude of the force that is applied to the rod is given by  

F = ilB

  = $\frac {B\omega a^2}{2R} \times a\times B$

  = $\frac {B^2\omega a^2}{2R}$