The magnetic field on the axis of a circular loop of radius 100 cm carrying current I = √(2) A, at point 1 m away from the centre of the loop is given by: (a) 6.28 × 10⁻⁴ T (b) 3.14 × 10⁻⁷ T (c) 6.28 × 10⁻⁷ T (d) 3.14 × 10⁻⁴ T

Answer: (b). B = μ₀ I R²2(R²+x²)3/2 with R = x = 1 m. B = 4π×10⁻⁷×√(2)× 12×(2)3/2 = 4π×10⁻⁷√(2)4√(2) = π×10⁻⁷ So the answer is (b) 3.14×10⁻⁷ T.

The magnetic field on the axis of a circular loop of radius 100 cm… | NEET 2022 (Phase 2)

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Biot-Savart Law & Circular Coils

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

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Answer: (b)

$B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ with $R = x = 1\text{ m}$ .
$B = \frac{4\pi\times10^{-7}\times\sqrt{2}\times 1}{2\times(2)^{3/2}} = \frac{4\pi\times10^{-7}\sqrt{2}}{4\sqrt{2}} = \pi\times10^{-7}$
So the answer is (b) $3.14\times10^{-7}\text{ T}$ .

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