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$B$.

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).

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.

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

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

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

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