An electron beam passes through a magnetic fi | vedclass

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Question

$\mathrm{B}=2 	imes 10^{-3}\, \mathrm{Wb} / \mathrm{m}^{2}$

$\mathrm{E}=1 	imes 10^{4}\, \mathrm{V} / \mathrm{m}^{2}$

Since the path of electr

Vedclass · Accepted Answer

$5	imes 10^6\,m/s,$ $1.43\,cm$

Vedclass · Answer

$\mathrm{B}=2 	imes 10^{-3}\, \mathrm{Wb} / \mathrm{m}^{2}$

$\mathrm{E}=1 	imes 10^{4}\, \mathrm{V} / \mathrm{m}^{2}$

Since the path of electron remains undeviated, $\mathrm{qvB}=\mathrm{q} \mathrm{E}$ or

$\mathrm{v}=\frac{\mathrm{E}}{\mathrm{B}}=\frac{1 	imes 10^{14}}{2 	imes 10^{-3}}=0.5 	imes 10^{7}$

$=5 	imes 10^{6}\, \mathrm{m} / \mathrm{s}$

If the electricfield is removed, the path of the charged particle is circular and magnetic field provides the necessary centripetal force.

i.e., $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bev} \Rightarrow \quad \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Be}}$

$=\frac{9.1 	imes 10^{-31} 	imes 5 	imes 10^{6}}{2 	imes 10^{-3} 	imes 1.6 	imes 10^{-19}}$

$=14.3 	imes 10^{-3}\, \mathrm{m}=1.43\, \mathrm{cm}$

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