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Question

The speed of charged particle just before entering the magnetic field is $\mathrm{V}_{0}$.

$\mathrm{qV}=\frac{1}{2} \mathrm{mv}_{0}^{2}$

$v_{0}=

Vedclass · Accepted Answer

$\frac{{q{B^2}{d^2}}}{{8V}}$

Vedclass · Answer

The speed of charged particle just before entering the magnetic field is $\mathrm{V}_{0}$.

$\mathrm{qV}=\frac{1}{2} \mathrm{mv}_{0}^{2}$

$v_{0}=\sqrt{\frac{2 q V}{m}}$

The radius of circular path in magnetic Field is

$r=\frac{m v_{0}}{q B}$

or $\frac{\mathrm{d}}{2}=\frac{\mathrm{mv}_{0}}{\mathrm{qB}}$

or $\frac{d^{2}}{4}=\frac{m^{2} v_{0}^{2}}{q^{2} B^{2}}=\frac{m^{2}}{q^{2} B^{2}} \frac{2 q V}{m}$

or $\frac{d^{2}}{4}=\frac{2 m V}{q B^{2}}$

$	herefore \quad \mathrm{m}=\frac{\mathrm{qB}^{2} \mathrm{d}^{2}}{8 \mathrm{V}}$

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