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 \(\vec{B} = 0.1\hat{k}\) Wb/m\(^2\). It moves a distance of \(\alpha\) meter along \(\hat{k}\) when it completes \(5\) revolutions. The value of \(\alpha\) is ________.

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: \(2\)