A conductor of length 2m, carrying a current of 10 A is kept in a magnetic field of induction `5xx10^(-4)Wb//m^(2).` The conductor experiences a force of `5xx10^(-3)N.` What is the angle made by the conductor, with the direction of the field ?

Options

A. `60^(@)`
B. `90^(@)`
C. `30^(@)`
D. `45^(@)`

Answer

C

Explanation

`F =5 xx10^(-3) N, B =5 xx 10^(-4) Wb//m^(2),`
`I =10 A, L =2m`
`F=Bil sin theta`
`therefore 5xx 10^(-3) =5 xx10^(-4) xx10xx2xx sin theta`
`therefore sin theta =(5xx10^(-3))/(5xx10^(-4) xx20) =1/2`
`therefore theta =30^(@)`