Question

A semicircular wire of radius r, carrying current i, is placed in a magnetic field at magnitude



b. The force acting on it

Answer 1

It is easy to see a pair of forces equal in magnitude and opposite in direction through RHL. So forces cancel out to zero.

But Calculating torque is critical. It forms a couple. Divide wire into a small segment with length dl and integrating whole from 0 to pi.

The angle between Force and dl is 90- @

dl = Rd@

so dT in the small segment = I dl B sin (90- @) x 2R cos @

= I dl cos @ 2r cos @

Integrating it under the limit 0- 90

You would get I (*pi* R^2)/2 * B

Answer 2

✓Hay mate✓

For semicircula ring magnetic field

B = (mu nod I / 4πR) × (theta)

Here,

Theta =π

Hence,

B =mu nod I / 4R

Now force acting on it

F = BIL

Mu nod I /4R × I × L

Where L(effective) = 2R

F = mu nod I^2 ×2R /4R

F = mu nod l^2/2

#phoenix