Question

A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T. The loop has an area of 0.14 m 2 and is mounted on an axis, perpendicular to the magnetic field, which allows the loop to rotate. What is the torque on the loop when its plane is oriented at a 21° angle to the field?

Answer 1

Answer:

The torque on the loop is $2.4 \times 10^{-2}$ Nm

Explanation:

Given:

Current $I = 1.6$ A

Magnetic field $B = 0.30$ T

Area of loop $A = 0.14$ $m^{2}$

Angle between magnetic field and area vector $\theta =$ 21°

Form the formula of  torque in case of magnetic field,

 г $= MB \sin \theta$

Where $M =$ magnetic moment

  $M = IA$

 г $= IAB \sin 21$

 г $= 1.6 \times 0.30 \times 0.14 \times 0.3583$

 г $=2.4 \times 10^{-2}$ Nm

Therefore, the torque on the loop is $2.4 \times 10^{-2}$ Nm