A charged particle of mass m and charge q moves with a constant velocity \( \vec{v}=a\hat{\imath} \) along the positive x-direction. From \( x=l \) it enters a region of uniform magnetic field \( \vec{B}=b\hat{k} \), directed along the positive z-direction, and moves in a circular path.
Find the radius of the circular path in which the particle moves.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \frac{ma}{qb} \)
Inside the field the magnetic force \( q\,\vec{v}\times\vec{B} \) is always perpendicular to the velocity. It changes only the direction of motion and acts as the centripetal force of a circular path.
Here \( \vec{v}=a\hat{\imath} \) is perpendicular to \( \vec{B}=b\hat{k} \), so the speed is \( v=a \) and the force has magnitude \( qvB \). Equating it to the centripetal force:
\( \displaystyle \frac{mv^{2}}{r}=qvB \)
\( \begin{aligned} r &= \frac{mv}{qB} \\ &= \frac{ma}{qb} \end{aligned} \)