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MHT CET 2025 (20 Apr, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2025 (20 Apr, Shift 2)

Q21MHT CET 2025 · 20 APR · SHIFT 2Medium

Biot-Savart Law & Circular Coils

An element Δl→=Δx i^\overrightarrow{\Delta l} = \Delta x\, \hat{i} is placed at the origin and carries a current of 10 A. The magnitude of magnetic field on the Y axis at a distance of 0.5 m if Δx=1 cm\Delta x = 1\text{ cm} is (μ04π=10−7\left(\frac{\mu_0}{4\pi} = 10^{-7}\right. SI unit ) (sin⁡90∘=1))\ (\sin 90^\circ = 1)

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Answer: (c)

  1. Biot-Savart law: dB=μ04πI Δxsin⁡θr2dB = \frac{\mu_0}{4\pi}\frac{I\,\Delta x \sin\theta}{r^2}, with θ=90∘\theta = 90^\circ for a point on the y-axis.
  2. dB=10−7×10×0.01(0.5)2=10−7×0.4dB = 10^{-7} \times \frac{10 \times 0.01}{(0.5)^2} = 10^{-7} \times 0.4
  3. So the answer is (c) 4×10−8 T4 \times 10^{-8}\text{ T}.
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