Question

A current carrying loop is free to turn in a uniform magnetic field. The loop will then come into equilibrium when it's plane is inclined at

• A. $0^0$ to the direction of the field.
• B. ${45}^0$ to the direction of the field.
• C. ${90}^0$ to the direction of the field.
• D. ${60}^0$to the direction of the field.

Answer 1

Answer: (C)

${90}^0$ to the direction of the field.

loop will come to equilibrium when torque is zero
So, $\overrightarrow{\tau }=0=\overrightarrow{M}\times \overrightarrow{B}$
$=MBsin\theta$
$=NiABsin\theta$
So, angle between area vector and field is zero than only $\theta$ is 0 so that $sin\theta =0$.
So, for area vector along field direction plane is perpendicular to the field.