Question

A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes

• A. inclined at 45$^{\circ}$ to the magnetic field
• B. inclined at any arbitrary angle to the magnetic field
• C. parallel to the magnetic field
• D. perpendicular to magnetic field

Answer 1

The Correct Option is D

To understand how a current-carrying coil behaves when subjected to a uniform magnetic field, we need to consider the physics of magnetic forces and torques. The question asks how the plane of the coil orients itself relative to the external magnetic field.

When a current-carrying coil is placed in a magnetic field, it experiences a torque due to the force exerted by the magnetic field on the moving charges within the coil. The torque (\(\tau\)) on a current loop in a magnetic field can be expressed as:

\(\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)\)

Where:

• n is the number of turns in the coil.
• I is the current through the coil.
• A is the area of the coil.
• B is the magnetic field strength.
• \(\theta\) is the angle between the normal to the plane of the coil and the magnetic field.

The coil will experience the maximum torque when \(\sin(\theta)\) is maximum, which happens when \(\theta = 90^{\circ}\), i.e., when the plane of the coil is perpendicular to the magnetic field (and the normal is parallel to the field). This orientation is energetically favorable because it minimizes the potential energy of the system.

On the other hand, if the plane of the coil is parallel to the magnetic field, the torque will be zero because \(\sin(0^{\circ}) = 0\). However, the coil will naturally orient itself to reach equilibrium in response to the torque, hence when the plane becomes perpendicular to the field, the system achieves this stable state.

Therefore, the correct answer is that the plane of the coil becomes perpendicular to the magnetic field.