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JEE Main 2023 (10 Apr, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2023 (10 Apr, Shift 2)

Q49JEE Main 2023 · 10 APR · SHIFT 2Easy

Biot-Savart Law & Circular Coils

A straight wire carrying a current of 14 A14\text{ A} is bent into a semi-circular arc of radius 2.2 cm2.2\text{ cm} as shown in the figure. The magnetic field produced by the current at the centre (O)(O) of the arc. is ____________ ×10−4 T\times 10^{-4}\text{ T}

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. The straight parts point along lines through O, so dl⃗×r⃗=0d\vec{l}\times\vec{r} = 0: they give no field at O.
  2. Semicircle: B=μ0I4R=4π×10−7×144×0.022B = \frac{\mu_0 I}{4R} = \frac{4\pi\times10^{-7}\times 14}{4\times 0.022}
  3. B=π×14×10−70.022=44×10−70.022=2×10−4 TB = \frac{\pi\times14\times10^{-7}}{0.022} = \frac{44\times10^{-7}}{0.022} = 2\times10^{-4}\text{ T}
  4. So the answer is 2 (i.e. 2×10−4 T2\times10^{-4}\text{ T}).
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