91Ó°ÊÓ

$$\begin{gathered} •Pitchofthepath6.00\mathrm{μ}m=6.00×{10}^{-6}m. \\ •MagneticfieldB=0.300T. \\ •Magneticforceontheelectron{F}_B=2.00×{10}^{-15}N. \end{gathered}$$

$$\begin{gathered} \mathbf{We}\mathbf{}\mathbf{use}\mathbf{}\mathbf{the}\mathbf{}\mathbf{formula}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{period}\mathbf{}\mathbf{of}\mathbf{}\mathbf{motion}\mathbf{}\mathbf{of}\mathbf{}\mathbf{charged}\mathbf{}\mathbf{particles} \\ \mathbf{}\mathbf{into}\mathbf{}\mathbf{the}\mathbf{}\mathbf{magnetic}\mathbf{}\mathbf{field}\mathbf{}\mathbf{into}\mathbf{}\mathbf{the}\mathbf{}\mathbf{formula}\mathbf{}\mathbf{of}\mathbf{}\mathbf{velocity}\mathbf{}\mathbf{to}\mathbf{}\mathbf{find}\mathbf{}\mathbf{the}\mathbf{}\mathbf{parallel}\mathbf{} \\ \mathbf{component}\mathbf{}\mathbf{of}\mathbf{}\mathbf{velocity}\mathbf{.}\mathbf{} \\ \mathbf{Then}\mathbf{}\mathbf{we}\mathbf{}\mathbf{use}\mathbf{}\mathbf{the}\mathbf{}\mathbf{formula}\mathbf{}\mathbf{of}\mathbf{}\mathbf{magnetic}\mathbf{}\mathbf{force}\mathbf{}\mathbf{to}\mathbf{}\mathbf{find}\mathbf{}\mathbf{the}\mathbf{} \\ \mathbf{perpendicular}\mathbf{}\mathbf{component}\mathbf{}\mathbf{of}\mathbf{}\mathbf{velocity}\mathbf{.}\mathbf{} \\ \mathbf{So}\mathbf{,}\mathbf{}\mathbf{by}\mathbf{}\mathbf{taking}\mathbf{}\mathbf{the}\mathbf{}\mathbf{magnitude}\mathbf{}\mathbf{of}\mathbf{}\mathbf{both}\mathbf{}\mathbf{components}\mathbf{}\mathbf{of}\mathbf{}\mathbf{velocity}\mathbf{,} \\ \mathbf{}\mathbf{we}\mathbf{}\mathbf{can}\mathbf{}\mathbf{find}\mathbf{}\mathbf{the}\mathbf{}\mathbf{electron}\mathbf{’}\mathbf{s}\mathbf{}\mathbf{speed}\mathbf{.}\mathbf{} \\ \mathrm{Formula}: \\ \mathrm{T}=2{\mathrm{Ï€³¾}}_{\mathrm{e}}/\mathrm{qB} \\ {\mathrm{F}}_{\mathrm{B}}=\mathrm{qVB} \\ \mathrm{V}=\mathrm{D}/\mathrm{T} \end{gathered}$$

$$\begin{gathered} Dis\tan cetraveledbytheelectronparalleltothemagneticfieldis: \\ {V}_{para}=\frac{D}{T} \\ But, \\ T=\frac{2\mathrm{π}m}{qB} \\ Theparallelcomponentofvelocitycanbecalculatedas: \\ {V}_{para}=\frac{D}{{\displaystyle \frac{2\mathrm{π}{m}_e}{qB}}} \\ =\frac{DqB}{2\mathrm{π}{m}_e} \\ =\frac{\left(6.00×{10}^{-6}m\right)\left(1.6×{10}^{-19}C\right)\left(0.300T\right)}{2\mathrm{π}\left(9.1×{10}^{-31}kg\right)} \\ =50395.4\frac{m}{s} \\ Now,themagneticforceisgivenby: \\ {F}_B=qB{V}_{per} \\ Theperpendicularcomponentofvelocitycanbecalculatedas: \\ {V}_{per}=\frac{F_B}{qB} \\ =\frac{\left(2.00×{10}^{-15}N\right)}{\left(1.6×{10}^{-19}C\right)\left(0.300T\right)} \\ =41666.7\frac{m}{s} \\ So,thespeedoftheelectronis: \\ V=\sqrt{{\left(V_{para}\right)}^2+{\left(V_{per}\right)}^2} \\ =\sqrt{{\left(50395.4\frac{m}{s}\right)}^2+{\left(41666.7\frac{m}{s}\right)}^2} \\ =65389.7\frac{m}{s} \\ =6.54×{10}^{4}m/s. \\ Therefore,theelectron’sspeedis6.54×{10}^{4}m/s. \end{gathered}$$