Question

A beam of protons with a velocity of $4\times {10}^5\text{m}/\text{s}$ enters a uniform magnetic field of $B=0.3\text{T}$. The velocity makes an angle of ${60}^{\circ }$ with the magnetic field. Find the pitch of the helix ?

• A. $7.5\text{cm}$
• B. $8.6\text{cm}$
• C. $3.7\text{cm}$
• D. $4.3\text{cm}$

Answer 1

The correct option is D $4.3\text{cm}$

Pitch is distance moved along field during one time period.

$\therefore$ Pitch$=V_{\parallel }\times T$

Velocity component of proton along the magnetic field is,

$V_{\parallel }=V\times \cos {60}^{\circ }=4\times {10}^5\times \left(\frac{1}{2}\right)$

$V_{\parallel }=2\times {10}^5\text{m}/\text{s}$

Time period of the proton,

$T=\frac{2\pi m}{qB}$

So, pitch $P=V_{\parallel }\times T$

$\Rightarrow \text{Pitch}=2\times {10}^5\times \left(\frac{2\pi \times 1.67\times {10}^{-27}}{1.6\times {10}^{-19}\times 0.3}\right)$

$\therefore$ Pitch $=0.0437\text{m}\approx 4.37\text{cm}$

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