A 5 mg particle carrying a charge of 5π× 10^ -6 C is moving with...

Given:
Mass of the particle, $m = 5 \text{ mg} = 5 \times 10^{-6} \text{ kg}$
Charge, $q = 5\pi \times 10^{-6} \text{ C}$
Velocity, $\vec{v} = (3\hat{i} + 2\hat{k}) \times 10^{-2} \text{ m/s}$
Magnetic field, $\vec{B} = 0.1\hat{k} \text{ Wb/m}^2$

The velocity component parallel to the magnetic field is $v_{\parallel} = 2 \times 10^{-2} \text{ m/s}$.

The time period of one revolution is given by:
$T = \dfrac{2\pi m}{qB}$

Substituting the given values:
$T = \dfrac{2\pi \times 5 \times 10^{-6}}{5\pi \times 10^{-6} \times 0.1} = \dfrac{2}{0.1} = 20 \text{ s}$

The time taken to complete $5$ revolutions is:
$t = 5T = 5 \times 20 = 100 \text{ s}$

The distance moved along the $\hat{k}$ direction (pitch for $5$ revolutions) is:
$\alpha = v_{\parallel} \times t$
$\alpha = 2 \times 10^{-2} \times 100 = 2 \text{ m}$

Answer: $2$