In Figure 28-40, an electron with an initial kinetic energy of$\mathbf{4}\mathbf{.0}$keV enters region 1 at time t= 0. That region contains a uniform magnetic field directed into the page, with magnitude $\mathbf{0}\mathbf{.010}\mathbf{}\mathit{T}$. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of$\mathbf{25}\mathbf{.0}$cm. There is an electric potential difference ∆$\mathit{V}\mathbf{=}\mathbf{}\mathbf{2000}$V across the gap, with a polarity such that the electron’s speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude $\mathbf{0}\mathbf{.020}$T. The electron goes through a half-circle and then leaves region 2. Atwhat time tdoes it leave?

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

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:

$T=2\pi m_e/qB$

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

$F=qE$

$F_B=qVB$

Time spent in region 1 is given by:

$\begin{array}{c}{t}_1=\frac{T_1}{2}\\ =\frac{2\pi m_e}{2q{B}_1}\\ =\frac{2\pi (9.1\times {10}^{-31}\text{\hspace{0.17em}kg})}{2(1.6\times {10}^{-19}\text{\hspace{0.17em}C})(0.01\text{\hspace{0.17em}T})}\\ =1.79\times {10}^{-9}\text{\hspace{0.17em}s}\end{array}$

Time spent in region 2 is given by:

$\begin{array}{c}{t}_2=\frac{T_2}{2}\\ =\frac{2\pi m_e}{2q{B}_2}\\ =\frac{2\pi (9.1\times {10}^{-31}\text{\hspace{0.17em}kg})}{2(1.6\times {10}^{-19}\text{\hspace{0.17em}C})\left(0.02\text{\hspace{0.17em}T}\right)}\\ =8.92\times {10}^{-10}\text{\hspace{0.17em}s}\end{array}$

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

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

Velocity can be calculated as:

$\begin{array}{c}6.4\times {10}^{-16}\text{\hspace{0.17em}J}=\frac{1}{2}(9.1\times {10}^{-31}\text{\hspace{0.17em}kg})v^2\\ v^2=\frac{(6.4\times {10}^{-16}\text{\hspace{0.17em}J})}{4.5\times {10}^{-31}\text{\hspace{0.17em}kg}}\\ v=37.5\times {10}^6\frac{\text{m}}{\text{s}}\end{array}$

And acceleration is:

$F=ma$

$\begin{array}{c}a=\frac{F}{m}\\ =\frac{eV}{m_{e}d}\\ =\frac{(1.6\times {10}^{-19}\text{\hspace{0.17em}J})(2000\text{V})}{(9.1\times {10}^{-31}\text{\hspace{0.17em}kg})(0.25\text{\hspace{0.17em}m})}\\ =1.41\times {10}^{15}\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

The time can be calculated as:

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

Solving this quadratic equation for time, we get:

$t_3=6.0\times {10}^{-9}s$

Therefore, the total time is:

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

Therefore, the time after which the electron leavesregion 2 is $8.7ns$.