JEE Challenger
More from Moving Charges and Magnetism
PhyMedAdv 2014 • Paper 2

Torque on Rotated Circular Loop in Field of Opposite Parallel Wires

Comprehension Passage

As illustrated in the diagram, a circular conducting loop of radius aa lies in the plane of the paper along with two long parallel wires (labeled 1 and 2). Both wires are situated at a distance dd from the center of the loop. Each wire as well as the circular loop carries an identical current II. Looking from above, the current circulating in the loop flows in the counterclockwise direction.

Passage Diagram 1

Assuming d≫ad \gg a, the circular loop is rotated through an angle of 30∘30^\circ about a diametrical axis parallel to the straight wires, starting from the configuration shown. If the currents through the two wires flow in mutually opposite directions, determine the torque acting on the loop in this new orientation (take the magnetic field generated by the wires to be uniform over the region of the loop).

Options

A

μ0I2a2d\frac{\mu_0 I^2 a^2}{d}

B

μ0I2a22d\frac{\mu_0 I^2 a^2}{2d}

Correct
C

3μ0I2a2d\frac{\sqrt{3}\mu_0 I^2 a^2}{d}

D

3μ0I2a22d\frac{\sqrt{3}\mu_0 I^2 a^2}{2d}

Topics & Concepts

Moving Charges and MagnetismBiot-Savart LawMagnetic Force

Step-by-Step Solution

To find the torque acting on the circular loop in its rotated orientation, we follow these steps:

1. Magnetic Field Due to the Two Wires

Let the plane of the paper be the xyxy-plane, with the center of the circular loop at the origin (0,0,0)(0,0,0). The two long parallel wires lie parallel to the yy-axis at positions x=−dx = -d (Wire 1) and x=+dx = +d (Wire 2).

Since the currents through the two wires flow in mutually opposite directions, let the current in Wire 1 be in the +y+y-direction (I⃗1=Ij^\vec{I}_1 = I\hat{j}) and the current in Wire 2 be in the −y-y-direction (I⃗2=−Ij^\vec{I}_2 = -I\hat{j}).

Using the Biot-Savart law (or Ampere's law) for an infinitely long straight wire, the magnetic field produced by each wire at the origin is:

  • Due to Wire 1 at x=−dx = -d: B⃗1=μ0I2πd(−k^)\vec{B}_1 = \frac{\mu_0 I}{2\pi d}(-\hat{k})
  • Due to Wire 2 at x=+dx = +d: B⃗2=μ0I2πd(−k^)\vec{B}_2 = \frac{\mu_0 I}{2\pi d}(-\hat{k})

Since d≫ad \gg a, the magnetic field over the region of the loop can be considered uniform and is given by the superposition of the fields from both wires: B⃗=B⃗1+B⃗2=−μ0Iπdk^\vec{B} = \vec{B}_1 + \vec{B}_2 = -\frac{\mu_0 I}{\pi d}\hat{k}

The magnitude of the net magnetic field is: B=μ0IπdB = \frac{\mu_0 I}{\pi d}

2. Magnetic Dipole Moment of the Loop

The circular loop carries a current II in the counterclockwise direction when viewed from above (along the +z+z-axis), meaning its initial magnetic dipole moment vector points perpendicular to the plane of the paper along the +z+z-direction: M⃗=IAk^=I(πa2)k^\vec{M} = I A \hat{k} = I (\pi a^2)\hat{k}

The magnitude of the magnetic moment is: M=Iπa2M = I \pi a^2

3. Torque on the Rotated Loop

The loop is rotated through an angle of θ=30∘\theta = 30^\circ about a diametrical axis parallel to the straight wires (the yy-axis).

Upon this rotation, the magnetic moment vector M⃗\vec{M} rotates by an angle of 30∘30^\circ away from the zz-axis. Since the magnetic field B⃗\vec{B} is oriented along the zz-axis, the angle between the magnetic moment vector M⃗\vec{M} and the magnetic field B⃗\vec{B} becomes θ=30∘\theta = 30^\circ (or 150∘150^\circ).

The magnitude of the torque τ\tau acting on the loop is: τ=∣M⃗×B⃗∣=MBsin⁡(30∘)\tau = |\vec{M} \times \vec{B}| = M B \sin(30^\circ)

Substituting the values of MM and BB: τ=(Iπa2)(μ0Iπd)sin⁡(30∘)\tau = (I \pi a^2) \left(\frac{\mu_0 I}{\pi d}\right) \sin(30^\circ)

Simplifying the expression: τ=(μ0I2a2d)(12)=μ0I2a22d\tau = \left(\frac{\mu_0 I^2 a^2}{d}\right) \left(\frac{1}{2}\right) = \frac{\mu_0 I^2 a^2}{2d}

Thus, the torque acting on the loop is μ0I2a22d\dfrac{\mu_0 I^2 a^2}{2d}.

Correct Option: (B)

PreviousNext