Question

A beam of protons with speed $4\times {10}^5$${\text{ms}}^{–1}$ enters a uniform magnetic field of $0.3\text{T}$ at an angle of ${60}^{\circ }$ to the magnetic field. The pitch of the resulting helical path of protons is close to :
$(\text{Mass of the proton}=1.67\times {10}^{–27}\text{kg},\text{charge of the proton}=1.6\times {10}^{–19}\text{C})$

• A. $2\text{cm}$
• B. $6\text{cm}$
• C. $12\text{cm}$
• D. $4\text{cm}$

Answer 1

The correct option is D $4\text{cm}$

When a charged particle enters a magnetic field making an angle $\theta$ with the magnetic field, then,

$\text{Pitch}=\left(v\cos \theta \right)T\text{and}T=\frac{2\pi m}{qB}$

$\therefore \text{Pitch}=\left(v\cos \theta \right)\frac{2\pi m}{qB}$

$\Rightarrow \text{Pitch}=(4\times {10}^5\cos {60}^{\circ })\frac{2\pi }{0.3}\left(\frac{1.67\times {10}^{-27}}{1.69\times {10}^{-19}}\right)$

$\Rightarrow \text{Pitch}\approx 4\times {10}^{-2}\text{m}=4\text{cm}$.

Hence, option $\left(D\right)$ is correct.