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JEE Main 2022 (24 Jun, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2022 (24 Jun, Shift 1)

Q73JEE Main 2022 · 24 JUN · SHIFT 1Medium

Biot-Savart Law & Circular Coils

The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance r2\frac{r}{2} from the centre is:

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Answer: (c)

  1. Baxis=Br3(r2+x2)3/2B_{axis} = B\dfrac{r^3}{(r^2+x^2)^{3/2}} with x=r2x = \dfrac{r}{2}.
  2. r2+x2=54r2⇒(r2+x2)3/2=(52)3r3r^2+x^2 = \dfrac{5}{4}r^2 \Rightarrow (r^2+x^2)^{3/2} = \left(\dfrac{\sqrt{5}}{2}\right)^3 r^3
  3. Baxis=(25)3BB_{axis} = \left(\dfrac{2}{\sqrt{5}}\right)^3 B
  4. So the answer is (c).
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