Question

Ask Your Teacher A 47.5-turn circular coil of radius 5.45 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.495 T. If the coil carries a current of 22.5 mA, find the magnitude of the maximum possible torque exerted on the coil.

Answer 1

Answer:

$4.9\times 10^{-3} Nm$

Explanation:

We are given that

Radius of coil=r=5.45 cm=$5.45\times 10^{-2} m$

$1 m=100 cm$

Number of turns,N=47.5

Magnetic field=B=0.495 T

Current,I=22.5 mA=$22.5\times 10^{-3} A$

$1m A=10^{-3} A$

We have to find the magnitude of maximum possible torque exerted on the coil.

We know that

Maximum torque,$\tau=NIAB$

Using the formula

$\tau=47.5\times 22.5\times 10^{-3}\times (\pi \times (5.45\times 10^{-2})^2\times 0.495$

$\tau=4.9\times 10^{-3} Nm$

Answer 2

Answer:

Explanation:

number of turns, N = 47.5

radius, r = 5.45 cm = 0.0545 m

Magnetic field, B = 0.495 T

current, i = 22.5 mA

Let the torque is τ.

τ = N x i x A x B x Sin 90

Angle is 90 for maximum torque.

τ = 47.5 x 0.0225 x π x 0.0545 x 0.0545 x 0.495

τ = 4.9 x 10^-3 Nm