JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 19)
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
).
7 × 10–6 C
4 × 10–5 C
2 × 10–6 C
3 × 10–5 C
Explanation
$$B = {{{\mu _0}i} \over {2a}}{{\omega q} \over {2\pi }} = i$$
$$B = {{{\mu _0}} \over {2a}}.{{\omega q} \over {2\pi }}$$
$$B = {{{{10}^{ - 7}} \times 40} \over {0.1}} \times q \times \pi $$
$$ \Rightarrow q = 3 \times {10^{ - 5}}C$$
$$B = {{{\mu _0}} \over {2a}}.{{\omega q} \over {2\pi }}$$
$$B = {{{{10}^{ - 7}} \times 40} \over {0.1}} \times q \times \pi $$
$$ \Rightarrow q = 3 \times {10^{ - 5}}C$$
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