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JEE Main 2024 (6 Apr, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2024 (6 Apr, Shift 1)

Q40JEE Main 2024 · 6 APR · SHIFT 1Medium

Biot-Savart Law & Circular Coils

An element Δl=Δx i^\Delta l = \Delta x\, \hat{i} is placed at the origin and carries a large current I=10 AI = 10\text{ A}. The magnetic field on the yy-axis at a distance of 0.5 m0.5\text{ m} from the elements Δx\Delta x of 1 cm1\text{ cm} length is:

Figure

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Answer: (c)

  1. Biot-Savart law: dB=μ04πI Δxsin⁡90∘r2dB = \frac{\mu_0}{4\pi}\frac{I\,\Delta x \sin 90^\circ}{r^2}.
  2. dB=10−7×10×10−20.25=4×10−8 TdB = 10^{-7} \times \frac{10 \times 10^{-2}}{0.25} = 4 \times 10^{-8}\text{ T}
  3. So the answer is (c) 4×10−8 T4 \times 10^{-8}\text{ T}.
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