91影视

We are given that length of parallel plates is $2.5cm$and distance between them (width) is $d=5.0mm$. An external $B=2.0mT$magnetic field of length $2.5cm$is applied perpendicular to the electric field between the plates.

Potential difference between the plates $鈭�V=600V$

Electron charge $q=1.6x{10}^{-19}C$

Electron mass role="math" localid="1649754340184" $m=9.1x{10}^{-31}Kg$

Since given in the question the electron came out undeflected so the electrical forces $F_e$equally balance the magnetic forces $F_b$.

$F_b=qvB$

$F_e=\frac{鈻�Vq}{d}$

Finding the electron speed v:

$$\begin{gathered} F_B=F_E \\ qvB=\frac{螖Vq}{d} \\ v=\frac{螖V}{dB} \\ v=\frac{600V}{(5.0脳{10}^{-3}m)(2.0脳{10}^{-3}T)} \\ v=6.0脳{10}^{7}m/s \end{gathered}$$

Since radius of curvature is only effected by the magnetic field and there is no role of the electric field (potential difference set to zero).

$F_b=\frac{m{v}^2}{r}$ (where m and q are mass and charge of electron respectively)

$$\begin{gathered} qvB=\frac{m{v}^2}{r} \\ r=\frac{mv}{qB} \\ r=\frac{(9.1脳{10}^{-31}kg)(6.0脳{10}^{7}m/s)}{(1.6脳{10}^{-19}C)(2.0脳{10}^{-3}T)} \\ r=0.171m \\ r=17.1cm \end{gathered}$$