A particle having a charge 50 e is revolving in a circular path of radius 0.4 m with 1 r.p.s. The magnetic field produced at the centre of the circle is (μ₀ = 4π × 10⁻⁷ SI units and e = 1.6 × 10⁻¹⁹ C) (a) 10⁻⁷μ₀ (b) 10⁻¹⁰μ₀ (c) 10⁻¹⁴μ₀ (d) 10⁻¹⁷μ₀

Answer: (d). I = qf = 50 × 1.6 × 10⁻¹⁹ × 1 = 8 × 10⁻¹⁸ A. B = (μ₀ I)/(2r) = μ₀ × 8 × 10⁻¹⁸0.8 = 10⁻¹⁷μ₀ So the answer is (d) 10⁻¹⁷μ₀.

A particle having a charge 50 e is revolving in a circular path of… | MHT CET 2024 (3 May, Shift 1)

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Biot-Savart Law & Circular Coils

A particle having a charge 50 e is revolving in a circular path of radius 0.4 m with $1\text{ r.p.s}$ . The magnetic field produced at the centre of the circle is $(\mu_0 = 4\pi \times 10^{-7}$ SI units and $e = 1.6 \times 10^{-19}\text{ C}$ )

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Answer: (d)

$I = qf = 50 \times 1.6 \times 10^{-19} \times 1 = 8 \times 10^{-18}\text{ A}$ .
$B = \frac{\mu_0 I}{2r} = \frac{\mu_0 \times 8 \times 10^{-18}}{0.8} = 10^{-17}\mu_0$
So the answer is (d) $10^{-17}\mu_0$ .

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