Question

A charged particle is accelerated through a potential difference of $12\text{kV}$ and acquires a speed of $1.0\times {10}^6\text{m/s}$. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.

• A. $12\text{cm}$
• B. $14\text{cm}$
• C. $10\text{cm}$
• D. None of these

Answer 1

The correct option is A $12\text{cm}$

When a charged particle is entered perpendicularly in a magnetic field,
$\frac{m{v}^2}{r}=qvBorr=\frac{mv}{qB}.....\left(1\right)$
$\frac{1}{2}m{v}^2=qV$
or $\frac{m}{q}=\frac{2V}{v^2}$
From equation (1),
$r=\frac{2V}{v^2}\times \frac{v}{B}$
$=\frac{2V}{vB}$
$=\frac{2\times 12\times {10}^3}{1\times {10}^6\times 0.2}$
$=12\text{cm}$