Answer:
Magnetic torque on the coil is 0.47 N-m.
Explanation:
It is given that,
Radius of the circular loop, r = 0.5 m
Magnetic field, B = 0.3 T
Current in the loop, I = 2 A
We need to find the magnetic torque when the plane of the loop is perpendicular to the magnetic field. The magnetic torque is given by :
[tex]\tau=BIA\ sin\theta[/tex]
[tex]\theta=90^{\circ}[/tex]
[tex]\tau=BI\pi r^2[/tex]
[tex]\tau=0.3\times 2\times \pi \times (0.5)^2[/tex]
[tex]\tau=0.47\ N-m[/tex]
So, the magnetic torque on the coil is 0.47 N-m. Hence, this is the required solution.