Acircular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 a. it is placed in a magnetic field of 0.70 t with the plane of the coil making an angle of 30° with the magnetic field. what is the magnetic torque on the coil?

τ = 0.088 N.m

Step by step explanation:

The magnetic torque of a N-turns circular coil of wire that is carrying a current and placed in a magnetic field is given by the equation:

τ = NIABsenφ

Where N is the number of turns of the circular coil of wire, I is the current flowing in the coil, A is the area of the coil, B is the magnetic field and senφ s the angle between the axis of the coil and the direction of the magnetic field.

To calculate the magnetic torque on the coil of 200 turns and diameter 2.0cm carrying a current of 4.0A placed in a magnetic field of 0.70T with the plane of the coil making an angle of 30° with the magnetic field:

A = πr² = π(d/2)²= πd²/4

τ = (200turns)(4.0A)[π(2.0x10⁻²m)²/4](0.70T)(sen 30°)

τ = 0.088 N.m

nah.   i'm just sleepy.