A $5$ mg particle carrying a charge of $5\pi \times 10^{-6}$ C is moving with velocity of $(3\hat{i} + 2\hat{k}) \times 10^{-2}$ m/s in a region having magnetic field $\vec{B} = 0.1\hat{k}\ \text{Wb/m}^2$. It moves a distance of $\alpha$ meter along $\hat{k}$ when it completes $5$ revolutions. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 2
The velocity component along $\vec{B}$, $v_\parallel = 2 \times 10^{-2}$ m/s, is unchanged; the particle moves on a helix.
Period: $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi \times 5 \times 10^{-6}}{5\pi \times 10^{-6} \times 0.1} = 20$ s.
Distance along $\hat{k}$ in $5$ revolutions: $5 \times T \times v_\parallel = 5 \times 20 \times 0.02 = 2$ m.
Solution by Sreeraj P, M.Sc Physics