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JEE Main 2024 (31 Jan, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from JEE Main 2024 (31 Jan, Shift 2)

Q37JEE Main 2024 · 31 JAN · SHIFT 2Medium

Current Loop as a Magnetic Dipole

A uniform magnetic field of 2×10−3 T2 \times 10^{-3}\text{ T} acts along positive YY-direction. A rectangular loop of sides 20 cm20\text{ cm} and 10 cm10\text{ cm} with current of 5 A5\text{ A} is in Y−ZY - Z plane. The current is in anticlockwise sense with reference to negative XX axis. Magnitude and direction of the torque is:

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Answer: (a)

  1. Anticlockwise as seen from the −X-X side means m⃗\vec m points along −i^-\hat i.
  2. m=IA=5×(0.2×0.1)=0.1 A m2m = IA = 5 \times (0.2 \times 0.1) = 0.1\text{ A m}^2, so m⃗=−0.1i^\vec m = -0.1\hat i.
  3. τ⃗=m⃗×B⃗=−0.1×2×10−3(i^×j^)=−2×10−4k^\vec\tau = \vec m \times \vec B = -0.1 \times 2\times10^{-3}(\hat i \times \hat j) = -2\times10^{-4}\hat k
  4. So the answer is (a) 2×10−42\times10^{-4} N m along negative ZZ.
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