The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction `1.5 xx 10^-2 T.` A current of 10.0 ampere flows through the loop. If the side of length 0.08 m is normal and the side of length 0.05 m is parallel to the lines of induction, then the torque acting on the loop is

Options

A. `6000 Nm`
B. zero
C. `1.2 xx 10^-2 Nm`
D. `6 xx 10^-4 Nm`

Answer

d

Explanation

Torque `tau` acting on a current carrying coil of area A placed
in a magnetic field of induction B is given by
`tau=NIBA sin theta`
where I=current in the coil, `theta`=anlgle which the normal to
the plane of the coil makes with the lines of induction B.
Here, `N=1, B=1.5xx10^-2T`
`A=0.05xx0.08=40xx10^-4m^2`
`I=10.0amp, theta=90^@ =pi//2`
`tau=(1.5xx10^-2)(10.0)xx(1)(40xx10^-4)sin(pi//2)`
`=6xx10^-4Nm`