What uniform magnetic field applied perpendicular to a beam of electrons moving at \( 1.3\times10^{6}\,\mathrm{m/s} \), is required to make the electrons travel in a circular arc of radius 0.35 m?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.1\times10^{-5}\,\mathrm{T} \)
The field is perpendicular to the velocity, so the magnetic force supplies the centripetal force and the electrons move on a circle.
\( \displaystyle evB = \frac{mv^{2}}{r} \Rightarrow B = \frac{mv}{er} \)
Substituting \( m = 9.1\times10^{-31}\,\mathrm{kg} \) and \( e = 1.6\times10^{-19}\,\mathrm{C} \):
\( \displaystyle B = \frac{9.1\times10^{-31}\times1.3\times10^{6}}{1.6\times10^{-19}\times0.35} = \frac{1.18\times10^{-24}}{5.6\times10^{-20}}\,\mathrm{T} \)
\( B \approx 2.1\times10^{-5}\,\mathrm{T} \)