A 150 - g ball containing$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$ excess electrons is dropped into a 125 - m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

The term magnetic field may be defined as the area around the magnet behave like a magnet.

When ball enter in the magnetic field and exit from the shaft that total velocity can be calculated from the equation

${\mathrm{v}}^2={\mathrm{v}}_0^2-2\mathrm{gs}$

The displacement is d = - y because the ball fall from y to 0 and the ${\mathrm{v}}_0=0$

And the final velocity is $\mathrm{v}=\sqrt{2\mathrm{gy}}$than put given values

$$\begin{gathered} \mathrm{v}=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(125\mathrm{m}\right)} \\ \mathrm{v}=49.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, final velocity is 49.5m/s

And the charge over the ball is

$$\begin{gathered} \mathrm{q}=-\mathrm{ne} \\ \mathrm{q}=-\left(4.00\times {10}^8\right)\left(1.602\times {10}^{-19}\mathrm{C}\right) \\ \mathrm{q}=-6.408\times {10}^{-11}\mathrm{C} \end{gathered}$$

Hence, charge over the ball is$-6.408\times {10}^{-11}\mathrm{C}$ .

Now the force acting on the ball${\overrightarrow{\mathrm{F}}}_{\mathrm{B}}=\mathrm{q}\left(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}\right)$ the direction of velocity is vertically downward and the direction of magnetic field to west and charge is negative. The direction of force out of the page due to right hand thumb rule. And the is the angle between magnetic field and velocity so magnitude of the force is calculated as

$$\begin{gathered} F_{\mathrm{B}}=\left|\mathrm{q}\right|\mathrm{vBsin}\left(\mathrm{\theta }\right) \\ {\mathrm{F}}_{\mathrm{B}}=\left(6.408\times {10}^{-11}\mathrm{C}\right)\left(49.5\mathrm{m}/\mathrm{s}\right)\left(0.250\mathrm{T}\right)\sin \left(90°\right) \\ {\mathrm{F}}_{\mathrm{B}}=7.93\times {10}^{-10}\mathrm{N} \end{gathered}$$

Hence the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field is$7.93\times {10}^{-10}\mathrm{N}$ to the south direction.