91Ó°ÊÓ

This force is due to the magnetic field, so to find the magnitude we apply our formula.

$$\begin{gathered} \stackrel{â‡\P Ä}{F}=q\stackrel{â‡\P Ä}{v}×\stackrel{â‡\P Ä}{B} \\ F=\left|q\right|vB\sin \left(\mathrm{Ï•}\right) \end{gathered}$$

Where $F$is the force we are looking for, $q$is the charge, $v$is the velocity of our ball, and $\mathrm{Ï•}$is the angle between the velocity vector and the magnetic field.

We have everything except the velocity. The charge is the charge of an electron times the number of electrons. The angle is 90 degrees so the sin term drops out, and the magnetic field strength is given.

To find the velocity, we can use energy concepts. Initially the ball is at rest, so the only energy is the gravitational potential energy. At the bottom of the shaft this has all converted into kinetic energy. Thus

localid="1650564691539" $$\begin{gathered} mgh=\frac{1}{2}m{v}^2 \\ v=\sqrt{2gh} \end{gathered}$$

Putting it all together, we have

localid="1650564694724" $$\begin{gathered} F=\left|q\right|vB\sin \left(\mathrm{Ï•}\right) \\ F=\left|q\right|\sqrt{2gh}B \end{gathered}$$

Now we can plug in our numbers and solve.

localid="1650564699179" $$\begin{gathered} F=\left|(4.10×{10}^8)·(-1.60*{10}^{-19}\mathrm{C})\right|\sqrt{2·9.8\mathrm{m}/{\mathrm{s}}^2·110\mathrm{m}}·(0.300\mathrm{T}) \\ F=9.14×{10}^{-10}\mathrm{N} \end{gathered}$$

We find the direction by applying the right hand rule.

The velocity vector is pointed down, and the magnetic field is pointed to the west, so our thumb points to the north. These are electrons though, and the right hand rule is defined for positive charges, so we must flip the direction of the force meaning the force is pointed to the south.