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

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this