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A charge \(q\) moving in a circle of radius \(r\) metre makes \(n\) rev/s. Magnetic field at the centre of the circle is
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Verified Answer
The correct answer is:
\(\frac{2 \pi n q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)
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, \(B=\frac{\mu_0 I}{2 r}\)
\(\begin{aligned}
& =\frac{\mu_0 \cdot q n}{2 r}=\frac{\left(4 \pi \times 10^{-7}\right) q n}{2 r} \\
& =\frac{2 \pi n q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}
\end{aligned}\)
\(\begin{aligned}
I & =\frac{q}{T}=\frac{q}{\left(\frac{1}{n}\right)} \quad\left[\because T=\frac{1}{n}\right] \\
& =q n
\end{aligned}\)
\(\therefore\) Magnetic field at the centre, \(B=\frac{\mu_0 I}{2 r}\)
\(\begin{aligned}
& =\frac{\mu_0 \cdot q n}{2 r}=\frac{\left(4 \pi \times 10^{-7}\right) q n}{2 r} \\
& =\frac{2 \pi n q}{r} \times 10^{-7} \mathrm{NA}^{-1} \mathrm{~m}^{-1}
\end{aligned}\)
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