Question

The effects due to the interaction of a current-carrying loop with a magnetic field have many applications, some as common as the electric motor. This problem illustrates the basic principles of this interaction.

Consider a current I that flows in a plane rectangular current loop with height a = 4.00cm and horizontal sides b = 2.00cm. The loop is placed into a uniform magnetic field B?  in such a way that the sides of length a are perpendicular to B? and there is an angle? between the sides of length b andB?.

For parts A and B, the loop is initially positioned at?=30?

A) Assume that the current flowing into the loop is 0.500A . If the magnitude of the magnetic field is 0.300T , what is ?, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?

B) What happens to the loop when it reaches the position for which ?=90?, that is, when its horizontal sides of length b are perpendicular to B?.
- The direction of rotation changes because the net torque acting on the loop causes the loop to rotate in a clockwise direction.
- The net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise direction.
- The net torque acting on the loop is zero; therefore it stops rotating.
- The net force acting on the loop is zero, so the loop must be in equilibrium.

Answer 1

Height of the loop, \(a=0.04 \mathrm{~m}\) Horizontal side of the beam, \(b=0.02 \mathrm{~m}\) Angle, \(\theta=30^{\circ}\) Magnetic field, \(B=0.3 \mathrm{~T}\) Current, \(I=0.5 \mathrm{~A}\)

A)

Net torque acting on the current loop is given by,

$$ \tau=L A B \sin \theta^{\prime} $$

Where \(\theta^{\prime}\) is the angle between the magnetic field and the normal to the loop.

$$ \begin{aligned} \theta^{\prime} &=90^{\circ}-\theta \\ &=90^{\circ}-30^{\circ} \\ &=60^{\circ} \end{aligned} $$

Hence torque is given by,

$$ \begin{aligned} \tau &=0.5 \times 0.04 \times 0.02 \times 0.3 \times \sin 60^{\circ} \\ &=1.04 \times 10^{-4} \mathrm{Nm} \end{aligned} $$

B)

When the loop reaches that particular position, the net torque acting on the loop is zero. But it continues to rotate in the counterclockwise direction.