91Ó°ÊÓ

• Kinetic energy$KE=4.0\text{ }KeV$
• The gap between the two regions$d=25\text{ }cm=0.25\text{ }m$
• The potential difference between two regions∆$V=2000$V
• The magnitude of the magnetic field in region 1$B_1=0.010\text{ }T$
• The magnitude of the magnetic field in region 1$F_2=0.020\text{ }T$

We can find the time spent by the particle in region 1 and region 2 by using the formula of the period of particle circulating in the magnetic field. Then by using the second kinematic equation, we can find the time spent by the particle between two regions. After adding all those times, we can find the time required for an electron to leave region 2.

Formula:

$T=2\mathrm{Ï€}{m}_e/qB$

$d=v_{0}t+\frac{1}{2}a{t}^2$

$F=qE$

$F_B=qVB$

Time spent in region 1 is given by:

$\begin{array}{c}{t}_1=\frac{T_1}{2}\\ =\frac{2\mathrm{Ï€}{m}_e}{2q{B}_1}\\ =\frac{2\mathrm{Ï€}(9.1×{10}^{−31}\text{ k²µ})}{2(1.6×{10}^{−19}\text{ C})(0.01\text{â€Í¿})}\\ =1.79×{10}^{−9}\text{ s}\end{array}$

Time spent in region 2 is given by:

$\begin{array}{c}{t}_2=\frac{T_2}{2}\\ =\frac{2\mathrm{Ï€}{m}_e}{2q{B}_2}\\ =\frac{2\mathrm{Ï€}(9.1×{10}^{−31}\text{ k²µ})}{2(1.6×{10}^{−19}\text{ C})\left(0.02\text{â€Í¿}\right)}\\ =8.92×{10}^{−10}\text{ s}\end{array}$

Time spent between the two regions is given by

$d=v{t}_3+\frac{1}{2}a{t}_3^2$

But,

$KE=\frac{1}{2}m{v}^2$

And

$\begin{array}{c}KE=4.0KeV\\ =(4.0×{10}^3\text{ }eV)(1.6×{10}^{−19}\text{ }J)\\ =6.4×{10}^{−16}\text{ }J\end{array}$

Velocity can be calculated as:

$\begin{array}{c}6.4×{10}^{−16}\text{ J}=\frac{1}{2}(9.1×{10}^{−31}\text{ k²µ})v^2\\ v^2=\frac{(6.4×{10}^{−16}\text{ J})}{4.5×{10}^{−31}\text{ k²µ}}\\ v=37.5×{10}^6\text{ }\frac{\text{m}}{\text{s}}\end{array}$

And acceleration is:

$F=ma$

$\begin{array}{c}a=\frac{F}{m}\\ =\frac{eV}{m_{e}d}\\ =\frac{(1.6×{10}^{−19}\text{ J})(2000\text{ }\text{V})}{(9.1×{10}^{−31}\text{ k²µ})(0.25\text{″¾})}\\ =1.41×{10}^{15}\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

The time can be calculated as:

$0.25\text{ }m=(37.5×{10}^6\text{ }\frac{m}{s})t_3+\frac{1}{2}(1.41×{10}^{15}\text{ }\frac{m}{s^2})t_3^2$

Solving this quadratic equation for time, we get:

$t_3=6.0×{10}^{−9}\text{ }s$

Therefore, the total time is:

$\begin{array}{c}t=t_1+t_2+t_3\\ =(1.79×{10}^{−9}\text{ }s)+(8.92×{10}^{−10}\text{ }s)+(6.0×{10}^{−9}\text{ }s)\\ =8.7×{10}^{−9}\text{ }s\\ =8.7\text{ }ns\end{array}$

Therefore, the time after which the electron leavesregion 2 is $8.7\text{ }ns$.