Question

An electron is projected into a uniform magnetic field of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $3\text{T}$ and moves along a helical path of radius <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $1\text{cm}$ with pitch $2\pi \text{cm}$. Find the angle of projection of the electron with the magnetic filed.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

We know, radius of the circular path is,

$r=\frac{m{v}_{\perp }}{qB}=\frac{mv\sin \theta }{qB}.......\left(1\right)$

and Pitch (P) is,

Pitch $\left(P\right)=v_{\parallel }\times T=v\cos \theta \times T=v\cos \theta \times \left(\frac{2\pi m}{qB}\right)..........\left(2\right)$

On dividing $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{r}{p}=\frac{\tan \theta }{2\pi }$

$\Rightarrow \tan \theta =2\pi \times \frac{r}{p}=2\pi \times \frac{1}{2\pi }$

$\Rightarrow \tan \theta =1$

$\therefore \theta ={45}^{\circ }$

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

Why this question ? Tip : Ratio of pitch and radius of the helical path only depends on the angle of projection.