Two protons are projected perpendicular to two different uniform magnetic fields \( B_1 \) and \( B_2 \) with velocities \( v_1 \) and \( v_2 \) respectively, where \( v_1=2v_2 \). If the radii of their circular paths are equal, what is the ratio \( B_2:B_1 \)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 : 2
A charged particle moving perpendicular to a uniform magnetic field describes a circle, with the magnetic force \( qvB \) acting as the centripetal force \( mv^{2}/r \).
Equating the two forces gives the radius:
\( \displaystyle \frac{mv^{2}}{r}=qvB \)
\( \displaystyle \Rightarrow\; r=\frac{mv}{qB} \)
Both particles are protons, so \( m \) and \( q \) are the same for the two paths. With equal radii:
\( \displaystyle \frac{mv_1}{qB_1}=\frac{mv_2}{qB_2} \)
\( \begin{aligned} \frac{B_2}{B_1} &= \frac{v_2}{v_1} \\ &= \frac{v_2}{2v_2} \\ &= \frac{1}{2} \end{aligned} \)
Therefore, \( B_2:B_1=1:2 \).