Question

The plane of the circular current carrying loop having radius 1.52 cm and mass 2.5 g is making an angle of 25 with the uniform magnetic field having a magnitude of 4.6x10- t. The current passing through the loop is 3.4

a. Calculate the magnitude of the angular acceleration of the circular loop. (a) 71 rad/s?

Answer 1

Given info : radius of circular current carrying loop is 1.52 cm and mass is 2.5g. it is making an angle of 25° with the uniform magnetic field having a magnitude of 4.6 × 10¯³T. the current passing through the loop is 3.4 A.

To find : The magnitude of the angular acceleration of the circular loop is ..

solution : torque of circular loop = BiNAsinθ

⇒I × α = BiNA sinθ

here, I = moment of inertia of circular loop = mr²

= 2.5 × 10¯³ kg × (1.52 × 10¯²m)²

B = magnetic field = 4.6 × 10¯³T

i = current = 3.4 A

N = no of turns = 1

A = area = π(1.52 × 10¯²)²

θ = angle made between B and A = 90° - 25° = 75°

now, 2.5 × 10¯³ kg × (1.52 × 10¯²m)² × α = 4.6 × 10¯³T × 3.4A × 1 × π(1.52 × 10¯² m)²sin65°

⇒2.5 × α = 4.6 × 3.4 × 3.14 × 0.906

⇒α = 18 rad/s² (approx)

Therefore the angular acceleration of the circular loop is 18 rad/s²