Question

A circular wire loop of radius 14.1 cm carries a current of 2.34 A. It is placed so that the normal to its plane makes an angle of 49.2° with a uniform magnetic field of magnitude 12.6 T. (a) Calculate the magnitude of the magnetic dipole moment of the loop in amperes-square meters. (b) What is the magnitude of the torque acting on the loop?

Answer 1

(a)
`0.1460Am^2`

(b) 1.3927 Nm

Step-by-step explanation:

We have given that radius
`r=14.1cm=0.141m`

Current i = 2.34 A

Angle
`\Theta =49.2^(\circ)`

Uniform magnetic field B = 12.6 T

Area is given by
`A=\pi r^2=3.14* 0.141^2=0.0624m^2`

(A) We know that magnetic dipole moment is given by
`M=iA` , here i is current and A is area

So magnetic moment
`M=iA=2.34* 0.0624=0.1460Am^2`

(b) We know that torque is given by

`\tau =BIAsin\Theta`

So torque experienced by coil will be
`\tau =BIAsin\Theta =12.6* 2.34* 0.0624* sin49.2^(\circ)=1.3927Nm`