An electron moves with a velocity of \( 1200\,\mathrm{m\,s^{-1}} \) in a magnetic field of 0.4 T at an angle of \( 30^\circ \). What will be the approximate radius of its path? (Take e/m of electron = \( 1.76 \times 10^{11}\,\mathrm{C\,kg^{-1}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 8.6 \times 10^{-9}\,\mathrm{m} \)
The magnetic force acts only on the component of velocity perpendicular to the field. The parallel component carries the electron along the field, so the path is a helix, and its radius is set by \( v_{\perp} = v\sin\theta \) alone.
\( \displaystyle r = \frac{m\,v\sin\theta}{eB} = \frac{v\sin\theta}{(e/m)\,B} \)
\( \displaystyle v\sin\theta = 1200 \times \sin 30^\circ = 600\,\mathrm{m\,s^{-1}} \)
\( \displaystyle (e/m)\,B = 1.76\times10^{11} \times 0.4 = 7.04\times10^{10}\,\mathrm{s^{-1}} \)
\( \displaystyle r = \frac{600}{7.04\times10^{10}} = 8.5\times10^{-9}\,\mathrm{m} \approx 8.6\times10^{-9}\,\mathrm{m} \)