Calculate Distance Moved Along Magnetic Field for Given Revolutions

To find the distance $\alpha$ moved along the $\hat{k}$ direction in $5$ revolutions, we analyze the helical motion of the charged particle in the uniform magnetic field.

1. Given Data:

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

2. Decomposition of Velocity: The magnetic field is directed along the $z$-axis ($\hat{k}$).

• Component of velocity parallel to $\vec{B}$: $v_\parallel = 2 \times 10^{-2}\text{ m/s}$
• Component of velocity perpendicular to $\vec{B}$: $v_\perp = 3 \times 10^{-2}\text{ m/s}$

Since the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ has no component along the direction of $\vec{B}$, the particle moves along $\hat{k}$ at a constant speed $v_\parallel$.

3. Time Period of One Revolution: The time period $T$ for one circular revolution in the plane perpendicular to the magnetic field is given by: $T = \frac{2\pi m}{q B}$

Substituting the given values into the formula: $T = \frac{2\pi \times (5 \times 10^{-6}\text{ kg})}{(5\pi \times 10^{-6}\text{ C}) \times 0.1\text{ T}}$

$T = \frac{10\pi \times 10^{-6}}{0.5\pi \times 10^{-6}} = 20\text{ s}$

4. Total Time for 5 Revolutions: The total time $t$ taken to complete $N = 5$ revolutions is: $t = N \times T = 5 \times 20\text{ s} = 100\text{ s}$

5. Calculation of Distance $\alpha$: The distance moved along the magnetic field ($\hat{k}$ direction) in total time $t$ is: $\alpha = v_\parallel \times t$ $\alpha = (2 \times 10^{-2}\text{ m/s}) \times 100\text{ s} = 2\text{ meters}$

Thus, the value of $\alpha$ is $2$.