A circular coil (radius = 0.40 m)has 160 turns and is in a uniform magnetic field.When the orientation of the coil is varied through all possible positions,the maximum torque on the coil by magnetic forces is 0.16 N . m when the current in the coil is 4.0 mA.What is the magnitude of the magnetic field?
A)0.37 T
B)1.6 T
C)0.50 T
D)1.2 T
E)2.5 T

Question

Quizplus · Accepted Answer

The torque on a current-carrying coil in a magnetic field is given by:
τ = NIABsinθ
where
N = number of turns in the coil
I = current in the coil
A = area of the coil
B = magnetic field strength
θ = angle between the normal to the coil and the direction of the magnetic field

In this case, the maximum torque is given as 0.16 Nm, N = 160, A = πr^2 = 0.16π m^2, I = 4.0 mA = 0.004 A. The angle θ is 90 degrees for maximum torque. Substituting these values, we get:
0.16 = (160)(0.004)(π)(0.4)^2B
B = 0.50 T

Therefore, the correct choice is C) 0.50 T.

A Circular Coil (Radius = 0