Question

A coil of 60 turns and area 1.5 × 10⁻³ m³ carrying 2 A current lies in a vertical plane. It experiences a torque of 0.12 Nm when placed in a uniform horizontal magnetic field. The torque acting on the coil changes to 0.05 Nm after the coil is rotated about its diameter by 90°, in the magnetic field. Find the magnitude of the magnetic field.

Answer 1

Given: N = 60

A = 1.5 × 10⁻³ m³

I = 2 A

τ₁ = 0.12    ...(when orientation is at θ)

τ₂ = 0.05

When orientation is at (90 + θ)

Thus, using the formula for torque on the coil.

Applying Torque for Two Cases:

τ = NIAB sin θ

τ₁ = NIAB sin (θ)    ...(1)

τ₂ = NIAB sin(90 + θ)

= NIAB cos(θ)    ...(2)

Dividing both equations,

tan(θ) = `tau_1/tau_2` = 2.4

Thus, sin(θ) = 0.92

Putting in equation (1),

B = 0.724 T