Question

A charged particle having mass $\left(m\right)$ and charge $\left(q\right)$ is projected from the origin with velocity $v=v_0\hat{i}$ in a uniform magnetic field, $B=\frac{B_0}{2}\hat{i}+\frac{\sqrt{3}{B}_0}{2}\hat{j}$ pitch of the helical path described by the particle is ?

• A. $\frac{2\pi m{v}_0}{B_{0}q}$
• B. $\frac{\sqrt{3}\pi m{v}_0}{2B_{0}q}$
• C. $\frac{\pi m{v}_0}{B_{0}q}$
• D. $\frac{2\sqrt{3}\pi m{v}_0}{B_{0}q}$

Answer 1

The correct option is C $\frac{\pi m{v}_0}{B_{0}q}$

$\because \text{pitch}\left(P\right)=v_{\parallel }\times T=v\cos \theta T=V\cos \theta \times \left(\frac{2\pi m}{qB}\right).........\left(1\right)$

Given,

$B=\frac{B_0}{2}\hat{i}+\frac{\sqrt{3}{B}_0}{2}\hat{j}$

$\therefore \tan \theta =\frac{B_y}{B_x}=\sqrt{3}\therefore \theta ={60}^{\circ }$

From eq (1) we get,

$P=v_0\times \frac{1}{2}\times \frac{2\pi m}{qB}$

Now,$B=\sqrt{{\left(\frac{B_0}{2}\right)}^2+{\left(\frac{\sqrt{3}{B}_0}{2}\right)}^2}=B_0$

$\therefore P=\frac{v_0}{2}\times \frac{2\pi m}{q{B}_0}$

$\therefore P=\frac{\pi m{v}_0}{B_{0}q}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, c is the correct option.

Why this question ? Hint : First you need to calculate angle between velocity vector and magnetic field and find the magnitude of the given magnetic field.