A positively charged $(+ q)$ particle of mass $m$ has kinetic energy $K$ enters vertically downward in a horizontal field of magnetic induction $\overrightarrow B $ . The acceleration of the particle is :-
Medium
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$\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}$
Thus, $v=\sqrt{\frac{2 \mathrm{K}}{\mathrm{m}}}$
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$
$\mathrm{F}=\mathrm{qvB} \sin \theta$
$\overrightarrow{\mathrm{v}} \perp \overrightarrow{\mathrm{B}},$ then $\theta=90^{\circ}$
$\therefore $ $\mathrm{F}=\mathrm{qvB}=\mathrm{qB} \sqrt{\frac{2 \mathrm{K}}{\mathrm{m}}}$
$a=\frac{F}{m}=\frac{q B \sqrt{2 K}}{(m)^{3 / 2}}$
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