Question
A charged particle is accelerated through a potential difference of 12kV and acquires a speed of 1.0 × 106m s−1. It is then injected perpendicularly into a magnetic field of strength 0.2T. Find the radius of the circle described by it.
✓
Answer
$\text{V}=12\text{KV}$
$\text{E}=\frac{\text{qV}}{\text{ml}}$
Now, $\text{F}=\text{qE}=\frac{\text{qV}}{\text{l}}$
$\text{a}=\frac{\text{F}}{\text{m}}=\frac{\text{qv}}{\text{ml}}$
$\text{v}=1\times10^6\text{m/s}$
$\text{v}=\sqrt{2\times\frac{\text{qV}}{\text{ml}}\times\text{l}}=\sqrt{2\times\frac{\text{q}}{\text{m}}\times12\times10^3}$
$1\times10^6=\sqrt{2\times\frac{\text{q}}{\text{m}}\times12\times10^3}$
$\Rightarrow10^{12}=24\times10^3\times\frac{\text{q}}{\text{m}}$
$\Rightarrow\frac{\text{m}}{\text{q}}=\frac{24\times10^3}{10^{12}}=24\times10^{-9}$
$\Rightarrow\frac{\text{mV}}{\text{qB}}=\frac{24\times10^{-9}\times1\times10^6}{2\times10^{-1}}=12\times10^{-2}\text{m}=12\text{cm}$
$\text{E}=\frac{\text{qV}}{\text{ml}}$
Now, $\text{F}=\text{qE}=\frac{\text{qV}}{\text{l}}$
$\text{a}=\frac{\text{F}}{\text{m}}=\frac{\text{qv}}{\text{ml}}$
$\text{v}=1\times10^6\text{m/s}$
$\text{v}=\sqrt{2\times\frac{\text{qV}}{\text{ml}}\times\text{l}}=\sqrt{2\times\frac{\text{q}}{\text{m}}\times12\times10^3}$
$1\times10^6=\sqrt{2\times\frac{\text{q}}{\text{m}}\times12\times10^3}$
$\Rightarrow10^{12}=24\times10^3\times\frac{\text{q}}{\text{m}}$
$\Rightarrow\frac{\text{m}}{\text{q}}=\frac{24\times10^3}{10^{12}}=24\times10^{-9}$
$\Rightarrow\frac{\text{mV}}{\text{qB}}=\frac{24\times10^{-9}\times1\times10^6}{2\times10^{-1}}=12\times10^{-2}\text{m}=12\text{cm}$
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