What is the maximum force on an aluminum rod with a 0.100-µC charge that you pass between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s? In what direction is the force?

The charge on the aluminum rod is \({\bf{q = 0}}{\bf{.100}}\;{\bf{\mu C}}\left( {\frac{{{\bf{1}}{{\bf{0}}^{{\bf{ - 6}}}}\;{\bf{C}}}}{{{\bf{1}}\;{\bf{\mu C}}}}} \right) = 1 \times {\bf{1}}{{\bf{0}}^{{\bf{ - }}7}}\;{\bf{C}}\)

The magnetic field strength of the permanent magnet is $\mathbf{B}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{T}$

The speed of the aluminum rod is role="math" localid="1653827291522" $\mathbf{v}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}$

To determine the force exerted on a charged rod in a magnetic field, we will use the following equation,

$\mathbf{F}\mathbf{=}\mathbf{qvBsin}\left(\theta \right)$……………………..(1)

Where $\theta$is the angle between the velocity and magnetic field direction. To get the maximum force, the direction of the velocity must be perpendicular to the magnetic field, i.e., having an angle of ${\mathbf{90}}^{\mathbf{°}}$or${\mathbf{270}}^{\mathbf{°}}$ where role="math" localid="1653827526323" $\mathit{\theta }$equals to 1 or -1, i.e., maximum.

The direction of the force will be perpendicular to both the velocity and the magnetic field. So, substituting the values in expression (1) will give,

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{qvB} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(1\times {10}^{‐7}C\right)\mathbf{\times }\left(5m/s\right)\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{T} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{7}\mathbf{}}\mathbf{N} \end{gathered}$$

Therefore, the maximum force is ${}^{\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{7}}\mathbf{}\mathbf{N}}$in the direction perpendicular to both the velocity and the magnetic field.