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JEE Main 2020 (6 Sep, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from JEE Main 2020 (6 Sep, Shift 2)

Q64JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Current Loop as a Magnetic Dipole

A square loop of side 2aa and carrying current I is kept in xz plane with its centre at origin. A long wire carrying the same current I is placed parallel to z-axis and passing through point (0, b, 0), (b >> a). The magnitude of torque on the loop about z-axis will be:

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Answer: (a)

  1. Since b≫ab \gg a, the wire's field at the loop is nearly uniform: B=μ0I2πbB = \frac{\mu_0 I}{2\pi b}, along the xx-axis.
  2. Loop in the xz plane: m=I(2a)2=4a2Im = I(2a)^2 = 4a^2I along the yy-axis, so m⃗⊥B⃗\vec m \perp \vec B.
  3. τ=mB=4a2I⋅μ0I2πb=2μ0I2a2πb\tau = mB = 4a^2I\cdot\frac{\mu_0 I}{2\pi b} = \frac{2\mu_0 I^2a^2}{\pi b}
  4. So the answer is (a).
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