Question

A rectangular coil of wire of 100 turns and 10 x 15cmsize carrying a current of 2Amp. is in a
magnetic field of induction 2 x 10 wb/m. If the normal drawn to the plane of the coil makes an
angle 30° with the field, then the torque on the coil is
1) 3x10-N-
m 2 ) 3x10-N-
m 3 ) 313 x 10-N-m 4) 313x10-N-m​

Answer 1

*I think, it should be like this*

Given: N = 100 turns , Area of rectangular coil = l x b = 10 x 15 = 150cm²

            I = 2A

            B = 2 x 10T

            Ф = 30°

Solution: Torque = NIBAsinФ

                             = 100 x 2 x 10 x 150 x sin 30°

                             = 300000 x 1/2

                             = 150000Nm

Answer 2

Answer:

3×10^-5

Explanation:

Torque=NIAB sin theta

=100×2×(10×15×10^-6)×2×10^-3×1/2

=2×150×10^-7

=300×10^-7N/m

=3×10^-5N/m