Question

A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of $1.0\times {10}^{6}m{s}^{-1}.$ It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.

Answer 1

$V_1=12KV=12\times {10}^{3}V,$

$F=qE,$

$v=1\times {10}^{6}m{s}^{-1}$

$v^2=u^2+2as$

$orv=2\times q\times qV/ml\times 1$

$=2\times q/m\times 12\times {10}^3$

$or(1\times {10}^6{)}^2=2\times q/m\times 12\times {10}^3$

$or{10}^{12}=24\times {10}^3\times q/m$

$or\frac{m}{q}=24\times {10}^{-9}$

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

$and=\frac{24\times {10}^{-9}\times {10}^6}{0.2}$

$r=12\times {10}^{-2}m=12cm$