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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2014 · Paper 2 · Physics · Magnetic Effects of Current

Paragraph: The figure shows a circular loop of radius a with two...

Paragraph: The figure shows a circular loop of radius aa with two long parallel wires (numbered 11 and 22 ) all in the plane of the paper. The distance of each wire from the centre of the loop is dd. The loop and the wires are carrying the same current II. The current in the loop is in the counterclockwise direction if seen from above. Paragraph: The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2 ) all in the

Question: Consider d≫ad \gg a, and the loop is rotated about its diameter parallel to the wires by 30∘30^{\circ} from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

Options

  1. Aμ0I2a22d\dfrac{{\mu}_{0}{I}^{2}{a}^{2}}{2d}
  2. Bμ0I2a2d\dfrac{{\mu}_{0}{I}^{2}{a}^{2}}{d}
  3. C3μ0I2a2d\dfrac{\sqrt{3}{\mu}_{0}{I}^{2}{a}^{2}}{d}
  4. D3μ0I2d22d\dfrac{\sqrt{3}{\mu}_{0}{I}^{2}{d}^{2}}{2d}

Correct answer

A. μ0I2a22d\dfrac{{\mu}_{0}{I}^{2}{a}^{2}}{2d}

Step-by-step solution

τ=MBsinθτ = MBsin\theta

⇒τ=I×πa2×2×μ0I2\pidsin30o=μ0I2a22d⇒ τ = I \times \pi{a}^{2} \times 2 \times \dfrac{{\mu}_{0}I}{2\pid}sin {30}^{o} = \dfrac{{\mu}_{0}{I}^{2}{a}^{2}}{2d}

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