A charge \(q\) moving in a circle of radius \(r\) metre makes \(n\) rev/s. Magnetic field at the centre of…

A charge \(q\) moving in a circle of radius \(r\) metre makes \(n\) rev/s. Magnetic field at the centre of the circle is
  1. \(\frac{2 \pi q}{n r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
  2. \(\frac{2 \pi q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
  3. \(\frac{2 \pi n q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
  4. \(\frac{2 \pi \mathrm{q}}{r} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)

Solution

Current associated due to the movement of charge \(q\) in circular path of radius \(r\) is given as \(\begin{aligned} I & =\frac{q}{T}=\frac{q}{\left(\frac{1}{n}\right)} \quad\left[\because T=\frac{1}{n}\right] \\ & =q n \end{aligned}\)
\(\therefore\) Magnetic field at the centre, \(B=\frac{\mu_0 I}{2 r}\) \(\begin{aligned} & =\frac{\mu_0 \cdot q n}{2 r}=\frac{\left(4 \pi \times 10^{-7}\right) q n}{2 r} \\ & =\frac{2 \pi n q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1} \end{aligned}\)

Asked in: AP EAMCET 2020 (18 Sep Shift 2)

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