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

 

Answers (1)

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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