A charge particle enters a uniform magnetic field with velocity vector at angle of ${45}^{\circ }$ with the magnetic field. The pitch of the helical path followed by the particle is $\text{P}$. The radius of the helix will be

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

From $\left(1\right)$ we get,

$P=\frac{v}{\sqrt{2}}\times \frac{2\pi m}{qB}.......\left(2\right)$

Radius, $r=\frac{m{v}_{\perp }}{qB}=\frac{mv\sin {45}^{\circ }}{qB}=\frac{mv}{\sqrt{2}qB}.......\left(3\right)$

On dividing $\left(3\right)$ by $\left(4\right)$ we get,

$\frac{P}{r}=2\pi$

$\therefore r=\frac{P}{2\pi }$

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