A positively charged (+ q) particle of mass m | vedclass

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Question

$\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}$

Thus, $v=\sqrt{\frac{2 \mathrm{K}}{\mathrm{m}}}$

$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarro

Vedclass · Accepted Answer

$\frac{{qB\sqrt {2K} }}{{{{(m)}^{\frac{3}{2}}}}}$

Vedclass · Answer

$\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}$

Thus, $v=\sqrt{\frac{2 \mathrm{K}}{\mathrm{m}}}$

$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} 	imes \overrightarrow{\mathrm{B}})$

$\mathrm{F}=\mathrm{qvB} \sin 	heta$

$\overrightarrow{\mathrm{v}} \perp \overrightarrow{\mathrm{B}},$ then $	heta=90^{\circ}$

$	herefore $ $\mathrm{F}=\mathrm{qvB}=\mathrm{qB} \sqrt{\frac{2 \mathrm{K}}{\mathrm{m}}}$

$a=\frac{F}{m}=\frac{q B \sqrt{2 K}}{(m)^{3 / 2}}$

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 :-

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