Question

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of Q with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the value of Q in the coil , if the torque is 0.96N

Answer 1

Answer:

Magnitude of torque experienced by coil in the magnetic field is given by,

τ=nBIAsinθ=0.96 Nm

where A=l×l

Explanation:

Answer 2

Answer:

Magnitude of torque experienced by coil in the magnetic field is given by,

τ=nBIAsinθ=0.96 Nm

where A=l×l