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

Answer:

(a)  $0.1460Am^2$

(b) 1.3927 Nm

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 $A=\pi r^2=3.14\times 0.141^2=0.0624m^2$

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

So magnetic moment $M=iA=2.34\times 0.0624=0.1460Am^2$

(b) We know that torque is given by

$\tau =BIAsin\Theta$

So torque experienced by coil will be $\tau =BIAsin\Theta =12.6\times 2.34\times 0.0624\times sin49.2^{\circ}=1.3927Nm$