Question

An electron at point$A$inFig. E27.15has a speed$v_0$of$1.41\times {10}^6\mathrm{m}/\mathrm{s}$Find (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from$A$to$B$, and (b) the time required for the electron to move from$A$to$B$.

Short answer

1. $B=1.6\times {10}^{-4}\mathrm{T}\mathrm{into}\mathrm{the}\mathrm{page}$
   
2. $t=1.1\times {10}^{-7}\mathrm{s}$
   

Step-by-step solution

27.15 (a) - Direction

To find the direction we can apply the right hand rule as follows:

Define our initial velocity to be in the positive y direction, then our electron feels a force in the positive x direction. Thus our B field must be pointed into the page (recall the right hand rule is defined for positive charges).

27.15 (a) - Magnitude

To find the magnitude, we must use our formula for the magnetic force on a particle.

$$\begin{gathered} \stackrel{\rightharpoonup }{F}=q\stackrel{\rightharpoonup }{v}\times \stackrel{\rightharpoonup }{B} \\ F=\left|q\right|vB\sin \left(\varphi \right) \end{gathered}$$

The magnitude of the magnetic field is not changeing, so we can choose a point where it is most convenient to calculate the magnitude. Thus

$F=\left|q\right|v_{0}B$

recall from Newton's second law that

$F=ma$

Where in this case, the acceleration is the centripital acceleration, so

$a=\frac{{v_0}^2}{r}$

Putting it all together we get

$$\begin{gathered} F=\left|q\right|v_{0}B \\ ma=\left|q\right|v_{0}B \\ m\frac{{v_0}^2}{r}=\left|q\right|v_{0}B \\ B=\frac{m{v}_0}{\left|q\right|r} \end{gathered}$$

These are all known values.

charge of an electron: $q=-1.6\times {10}^{-19}\mathrm{C}$

mass of an electron: $m=9.11\times {10}^{-31}\mathrm{kg}$

The radius is half of our given diameter, so $r=5.0\mathrm{cm}=0.05\mathrm{m}$

Now we can plug in our values and evaluate.

$B=\frac{(9.11\times {10}^{-31}\mathrm{kg})·(1.41\times {10}^6\mathrm{m}/\mathrm{s})}{(1.60\times {10}^{-19}\mathrm{C})·(0.05\mathrm{m})}$

$$\begin{gathered} B=1.6*{10}^{-4}\frac{\mathrm{kg}}{\mathrm{C}·\mathrm{s}} \\ B=1.6\times {10}^{-4}\mathrm{T} \end{gathered}$$

27.15 (b) - time

To find the time we can use rotational motion. The period is the time to complete one full circle, so our time is half the period

$$\begin{gathered} t=\frac{1}{2}T \\ t=\frac{1}{2}\left(\frac{2\pi }{\omega }\right) \\ t=\frac{\mathrm{\pi }}{\mathrm{\omega }} \end{gathered}$$

Recall that our angular velocity, $\omega$, is given in terms of the tangential velocity as follows:

$\omega =\frac{v}{r}$

Thus we have

$$\begin{gathered} t=\frac{\pi }{\omega } \\ t=\frac{\pi }{\left({\displaystyle \frac{v_0}{r}}\right)} \\ t=\frac{\mathrm{\pi r}}{{\mathrm{v}}_0} \end{gathered}$$

Plug in our numbers and evaluate

$$\begin{gathered} t=\frac{\pi ·0.05\mathrm{m}}{1.41\times {10}^6\mathrm{m}/\mathrm{s}} \\ t=1.1\times {10}^{-7}\mathrm{s} \end{gathered}$$