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. `BI l`

Answer

C

Explanation

`BIl=2Tsintheta`
`theta` is small, `sintheta=theta`
`:. BIl=2Ttheta`

or `Bil=(2Tl//2)/(r )`
or `BI=(T)/(r )`
But `2pil=1` or `r=(l)/(2pi)`
` :. BI=(T2pi)/(l)` or `T=(Bil)/(2pi)`