Question

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
A. 6000 newton `xx` metre
B. zero newton `xx` metre
C. `1.2xx10^(-2)` newton `xx` metre
D. `6xx10^(-4)` newton `xx` metre

Answer 1

Correct Answer - D
Torque `tau` acting on a current carrying coil of area `A` placed in a magnitude field of induction `B` is given by:
`tau=NIBAsintheta`
Where `I=` current in the coil, `theta=` angle which the normal to the plane of the coil makes with the lines of induction `B`.
Here `N=1,B=1.5xx10^(-2)tesla,`
`A=0.05xx0.08=40xx10^(-4)m^(2),`
`I=10.0amp,=90^(@)=pi//2`
`tau=(1.5xx10^(-2))(10.0)xx(1)(40xx10^(-4))sin((pi)/(2))`
`=6xx10^(-4)` newton `xx` meter