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

The torque on the loop is
`2.4 * 10^(-2)` Nm

Step-by-step 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 * 0.30 * 0.14 * 0.3583`

г
`=2.4 * 10^(-2)` Nm

Therefore, the torque on the loop is
`2.4 * 10^(-2)` Nm