(a) A pendulum is set up so that its bob (a thin copper disk) swings between the poles of a permanent magnet as shown in Figure 22.63. What is the magnitude and direction of the magnetic force on the bob at the lowest point in its path, if it has a positive$\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{\mu C}$charge and is released from a height of 30.0 cmabove its lowest point? The magnetic field strength is 1.50 T (b) What is the acceleration of the bob at the bottom of its swing if its mass isgrams and it is hung from a flexible string? Be certain to include a free-body diagram as part of your analysis.

A magnetic field is defined as a position in space near a magnet or an electric current where a physical field is formed by a moving electric charge applying force on another moving electric charge.

Height is $30\mathrm{cm}\left(\frac{1\hspace{0.33em}\mathrm{m}}{100\mathrm{cm}}\right)=0.3\hspace{0.33em}\mathrm{m}$

Charge is $0.250\hspace{0.33em}\mathrm{\mu C}\left(\frac{{10}^{-6}\hspace{0.33em}\mathrm{C}}{1\mathrm{\mu C}}\right)=2.50\times {10}^{-7}\mathrm{C}$

Magnetic field strength is 1.50 T

$\mathrm{P}.\mathrm{E}=\mathrm{K}.\mathrm{E}\to \mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2$

(a)

We apply the equation$\mathrm{F}=\mathrm{qvB}$to find the amount of the magnetic force at the lowest position, but we must first compute the bob's velocity. When the bob reaches a certain height, it gains potential energy, which can be computed using the equation, and when the bob is released, all of the potential energy converts to kinetic energy at the lowest position, in other words, $\mathrm{P}.\mathrm{E}=\mathrm{K}.\mathrm{E}\to \mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2.$

We'll solve this equation for v and use the result to calculate the magnetic force magnitude.

$$\begin{gathered} \mathrm{v}=\sqrt{2\mathrm{gh}} \\ =\sqrt{2\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.3\mathrm{m}} \\ =2.43\mathrm{m}/\mathrm{s} \\ \mathrm{F}=\mathrm{qvB} \\ =2.50\times {10}^{-7}\mathrm{C}\times 2.43\mathrm{m}/\mathrm{s}\times 1.50\mathrm{T} \\ =9.1\times {10}^{-7}\mathrm{N} \end{gathered}$$

Therefore, magnetic force acting on the bob is $9.1\times {10}^{-7}\mathrm{N}$

(b)

The right-hand rule will be used to determine the force's direction. The palm will be pointed to the force's direction, which is upwards, by adjusting the thumb to the velocity's direction (to the right) and the fingers to the field's direction (into the page).

In the absence of a magnetic field, the bob has a maximum velocity at the lowest location, implying that the acceleration is zero. In this situation, just two forces are acting on the bob: its weight downwards, $\mathrm{Fg}\downarrow$, and the tension in the rope upwards, $\mathrm{T}\uparrow$, and the two forces are equal.

There is an acceleration in the presence of the magnetic field since the net force is no longer zero, and this acceleration is in the direction of the net force (upwards).

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=\mathrm{ma} \\ =\mathrm{F}+\mathrm{T}-\mathrm{Fg} \\ =\mathrm{F} \end{gathered}$$

localid="1656409640278" $$\begin{gathered} \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \\ =\frac{9.1\times {10}^{-7}\mathrm{N}}{30.0\times {10}^{-3}\mathrm{kg}} \\ =3.03\times {10}^{-5}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration at the lowest point is $3.03\times {10}^{-5}\mathrm{m}/{\mathrm{s}}^2$