A thin ring of 10 cm radius carries a uniformly distributed charge… | JEE Main 2019 (12 Apr, Shift 1)

Q89JEE Main 2019 · 12 APR · SHIFT 1Medium

Biot-Savart Law & Circular Coils

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 $\pi$ rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9 T, then the charge carried by the ring is close to ( $\mu$ 0 = 4 $\pi$ × 10–7 N/A2 ).

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Answer: (d)

$f = \frac{\omega}{2\pi} = 20\text{ Hz}$ , so $I = qf$ and $B = \frac{\mu_0 q f}{2R}$ .
$q = \frac{2RB}{\mu_0 f} = \frac{2 \times 0.1 \times 3.8 \times 10^{-9}}{4\pi \times 10^{-7} \times 20}$
$q = \frac{7.6 \times 10^{-10}}{2.51 \times 10^{-5}} \approx 3 \times 10^{-5}\text{ C}$
So the answer is (d) $3 \times 10^{-5}\text{ C}$ .

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