Question

A proton projected in a magnetic field of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $0.02\text{T}$ moves along a helical path of radius <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $5\text{cm}$ and pitch <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $20\text{cm}$. Find the componet of velocity parallel to the magnetic field. Take the mass of the proton $1.6\times {10}^{-27}\text{kg}$

• A. $6.4\times {10}^3\text{m/s}$
• B. $6.4\times {10}^4\text{m/s}$
• C. $3.2\times {10}^4\text{m/s}$
• D. $3.2\times {10}^3\text{m/s}$

Answer 1

The correct option is B $6.4\times {10}^4\text{m/s}$

Given, $B=0.02\text{T}r=5\text{cm}=5\times {10}^{-2}\text{m}\text{pitch}\left(P\right)=20\text{cm}=20\times {10}^{-2}\text{m}$

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

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

$2\times {10}^{-2}=\frac{v\cos \theta \times 2\pi \times 1.6\times {10}^{-27}}{1.6\times {10}^{-19}\times 0.2}$

$\Rightarrow v\cos \theta \approx 6.4\times {10}^4\text{m/s}$

$v\cos \theta$ is velocity along the magnetic filed.

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

Why this question? Tip : In the pitch formula, we use velocity component which is parallel to the magnetic field.