An electron makes a circular motion of radius 5x10−10 m around a proton with a speed 2x106 m/s. The magnetic field produced by the electron at the center will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.128 T
The revolving electron is a circular current loop: its charge passes any point once per revolution, so the current is e divided by the period. The field at the centre of a loop is \( \mu_0 I/2r \).
\( \displaystyle I = \frac{e}{T} = \frac{ev}{2\pi r} \)
\( \displaystyle B = \frac{\mu_0 I}{2r} = \frac{\mu_0 e v}{4\pi r^{2}} \)
\( \displaystyle B = 10^{-7} \times \frac{1.6\times10^{-19} \times 2\times10^{6}}{(5\times10^{-10})^{2}} \)
\( \displaystyle B = 10^{-7} \times \frac{3.2\times10^{-13}}{2.5\times10^{-19}} \)
\( \displaystyle B = 0.128\,\mathrm{T} \)