Question

Derive an expression for torque acting on a rectangular current carrying loop kept in a uniform magnetic field B. Indicate the direction of torque acting on the loop.

OR

Deduce the expression for the torque \(\overrightarrow{t}\) acting on a planar loop of area \(\overrightarrow{A}\) and carrying current I placed in a uniform magnetic field \(\overrightarrow{B}\) If the loop is free to rotate, what would be its orientation in stable equilibrium?

Answer 1

Torque on a current carrying loop: Consider a rectangular loop PQRS of length l, breadth b suspended in a uniform magnetic field \(\overrightarrow{B}\) The length of loop = PQ = RS = l

and breadth QR = SP = b. Let at any instant the normal to the plane of loop make an angle θ with the direction of magnetic field \(\overrightarrow{B}\) and I be the current in the loop. We know

that a force acts on a current carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. Suppose that the forces on

sides PQ, QR, RS and SP are \(\overrightarrow{F_1}\), \(\overrightarrow{F_2}\), \(\overrightarrow{F_3}\) and \(\overrightarrow{F_4}\) respectively. The sides QR and SP make angle (90°– θ) with the direction of magnetic field. Therefore each of the forces \(\overrightarrow{F_2}\) and \(\overrightarrow{F_4}\) acting on these sides has same magnitude F′ = Blb sin (90°– θ) = Blb cos θ. According to Fleming’s left hand rule the forces are equal and opposite but their line of action is same. Therefore these forces cancel each other i.e. the resultant of \(\overrightarrow{F_2}\) and \(\overrightarrow{F_4}\) is zero.

The sides PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces \(\overrightarrow{F_1}\) and \(\overrightarrow{F_3}\)  acting on sides PQ and RS are equal and opposite, but their lines of action are different; therefore the resultant force of \(\overrightarrow{F_1}\) and \(\overrightarrow{F_3}\) is zero, but they form a couple called the deflecting couple. When the normal to plane of loop makes an angle with the direction of magnetic field the perpendicular distance between F₁ and F₃ is b sin θ.

∴ Moment of couple or Torque,

τ = (Magnitude of one force F) × perpendicular distance = (BIl). (b sin θ)=I (lb) B sin θ

But lb = area of loop = A (say)

∴ Torque, τ = IAB sin θ

If the loop contains N-turns, then τ = NI AB sin θ

The magnetic dipole moment of rectangular current loop = M = NIA

Direction of torque is perpendicular to direction of area of loop as well as the direction of magnetic field i.e., along \(I\overrightarrow{A}\times \overrightarrow{B}.\)

The current loop would be in stable equilibrium, if magnetic dipole moment is in the direction of the magnetic field (\(\overrightarrow{B}\)).