# 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 $3.8\times10^{-9}$ T, then the charge carried by the ring is clode to $(\mu_{0}=4\pi\times10^{-7}N/A^{2})$. Option 1) $7\times10^{-6}C$ Option 2) $3\times10^{-5}C$ Option 3) $2\times10^{-6}C$ Option 4) $4\times10^{-5}C$

We know that,

Magnetic field at the centre of a current carrying ring, $B=\frac{\mu_{0}i}{2R}$

$i=\frac{q}{T}$

$T=\frac{vw}{2\pi}$

$B=\frac{\mu_{0}qw}{2R.2\pi}$

$\Rightarrow q=\frac{4\pi RB}{\mu_{0}w}$

$\Rightarrow q=3\times10^{-5}C$

Option 1)

$7\times10^{-6}C$

Option 2)

$3\times10^{-5}C$

Option 3)

$2\times10^{-6}C$

Option 4)

$4\times10^{-5}C$

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