A charged particle with charge 0.01 C rotates on a circular path of radius 0.5 m, with velocity 0.2 m/s. What is the value of B at the center of the circle?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8\times10^{-10}\,\mathrm{T} \)
A charge going round a circle is an equivalent circular current \( I = q/T \), and the field at the centre of a circular loop of radius r is \( B = \frac{\mu_0 I}{2r} \).
Time for one revolution:
\( \displaystyle T = \frac{2\pi r}{v} = \frac{2\pi\times0.5\,\mathrm{m}}{0.2\,\mathrm{m\,s^{-1}}} = 5\pi\,\mathrm{s} \)
\( \displaystyle I = \frac{q}{T} = \frac{0.01\,\mathrm{C}}{5\pi\,\mathrm{s}} = \frac{0.002}{\pi}\,\mathrm{A} \)
\( \displaystyle B = \frac{\mu_0 I}{2r} = \frac{4\pi\times10^{-7}\times\frac{0.002}{\pi}}{2\times0.5} = 8\times10^{-10}\,\mathrm{T} \)
Shortcut: Putting \( I = \frac{qv}{2\pi r} \) into \( B = \frac{\mu_0 I}{2r} \) gives \( B = \frac{\mu_0}{4\pi}\frac{qv}{r^{2}} \) directly:
\( \displaystyle B = 10^{-7}\times\frac{0.01\times0.2}{(0.5)^{2}} = 8\times10^{-10}\,\mathrm{T} \)