A loop of flexible conducting wire of length l lies in magnetic field B which is normal to the plane of loop. A current l is passed through the loop. The tension developed in the wire to open up is

Options

A. `pi/2 Bil`
B. `(Bil)/2`
C. `(Bil)/(2pi)`
D. `Bil`

Answer

c

Explanation

`BI(dl)=2Tsin (d theta//2)`
`implies BI(r d theta)=2T(d theta//2)` ( `theta` is small, `sin theta= theta`)
`implies T=Bir=Bil//2pi`
.