Answer:
(a) [tex]0.1460Am^2[/tex]
(b) 1.3927 Nm
Explanation:
We have given that radius [tex]r=14.1cm=0.141m[/tex]
Current i = 2.34 A
Angle [tex]\Theta =49.2^{\circ}[/tex]
Uniform magnetic field B = 12.6 T
Area is given by [tex]A=\pi r^2=3.14\times 0.141^2=0.0624m^2[/tex]
(A) We know that magnetic dipole moment is given by [tex]M=iA[/tex] , here i is current and A is area
So magnetic moment [tex]M=iA=2.34\times 0.0624=0.1460Am^2[/tex]
(b) We know that torque is given by
[tex]\tau =BIAsin\Theta[/tex]
So torque experienced by coil will be [tex]\tau =BIAsin\Theta =12.6\times 2.34\times 0.0624\times sin49.2^{\circ}=1.3927Nm[/tex]
