Q131 / 200 · NEET 2022 (Phase 2) · Physics

The magnetic field on the axis of a circular loop of radius $100 \mathrm{~cm}$ carrying current $I=\sqrt{2} \mathrm{~A}$, at point $1 \mathrm{~m}$ away from the centre of the loop is given by:

  • Correct answer
    A
    $3.14 \times 10^{-7} \mathrm{~T}$
  • B
    $6.28 \times 10^{-7} \mathrm{~T}$
  • C
    $3.14 \times 10^{-4} \mathrm{~T}$
  • D
    $6.28 \times 10^{-4} \mathrm{~T}$

Solution

As per question, we have; $\begin{aligned} \mathrm{B} & =\mathrm{B}_0 \sin ^3 \theta \\ & =\frac{\mu_0 \mathrm{I}}{2 \pi} \sin ^3\left(45^{\circ}\right) \\ & =\frac{4 \pi \times 10^{-7} \times \sqrt{2}}{2 \times 1}\left(\frac{1}{\sqrt{2}}\right)^3 \\ & =3.14 \times 10^{-7} \mathrm{~T} \end{aligned}$

Hence, the correct answer is option: 'A'.

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