Question

Alpha particle ($m=6.7\times {10}^{-27}kg,q=+2e$) are accelerated from rest through a potential difference of 6.7 kV. Then, they enter a magnetic field B= 0.2 T perpendicular to them direction of their motion. The radius of the path described by them is

• A. $8.375m$
• B. 8.375 cm
• C. Infinity
• D. None of these

Answer 1

The correct option is B 8.375 cm

We know,

$R=\frac{mv}{qB}$

$R=\frac{\sqrt{2mqV}}{qB}$

$R=\frac{\sqrt{2mV}}{\sqrt{q}B}$

$R=\frac{\sqrt{2}\times 6.7\times {10}^{-12}}{\sqrt{2}\times 2\times {10}^{-10}\times 0.2}$

$R=\frac{1.675\times {10}^{-2}}{0.2}$

$R=8.375cm$

Option $\text{B}$ is the correct answer.