Question

$$\begin{gathered} \mathbf{A}\mathbf{}\mathbf{particle}\mathbf{}\mathbf{undergoes}\mathbf{}\mathbf{uniform}\mathbf{}\mathbf{circular}\mathbf{}\mathbf{motion}\mathbf{}\mathbf{of}\mathbf{}\mathbf{radius}\mathbf{}\mathbf{26}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{\mu }\mathbf{m}\mathbf{}\mathbf{in}\mathbf{}\mathbf{a}\mathbf{}\mathbf{uniform}\mathbf{}\mathbf{magnetic}\mathbf{}\mathbf{field}\mathbf{.}\mathbf{} \\ \mathbf{The}\mathbf{}\mathbf{magnetic}\mathbf{}\mathbf{force}\mathbf{}\mathbf{on}\mathbf{}\mathbf{the}\mathbf{}\mathbf{particle}\mathbf{}\mathbf{has}\mathbf{}\mathbf{a}\mathbf{}\mathbf{magnitude}\mathbf{}\mathbf{of}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathbf{}\mathbf{}\mathbf{N}\mathbf{.} \\ \mathbf{}\mathbf{What}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{kinetic}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{particle}\mathbf{?} \end{gathered}$$

Short answer

$\mathrm{The}\mathrm{kinetic}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{is}2.09\times {10}^{-22}\mathrm{J}.$

Step-by-step solution

Listing the given quantities

$$\begin{gathered} •\mathrm{Radius}\mathrm{r}=26.1\mathrm{\mu m}=26.1\times {10}^{-6}\mathrm{m}. \\ •\mathrm{Magnetic}\mathrm{force}\mathrm{F}=1.60\times {10}^{-17}\mathrm{N} \end{gathered}$$

Understanding the concept of kinetic energy and centripetal force

$$\begin{gathered} \mathbf{We}\mathbf{}\mathbf{use}\mathbf{}\mathbf{the}\mathbf{}\mathbf{formula}\mathbf{}\mathbf{of}\mathbf{}\mathbf{centripetal}\mathbf{}\mathbf{force}\mathbf{}\mathbf{to}\mathbf{}\mathbf{derive}\mathbf{}\mathbf{the}\mathbf{}\mathbf{formula}\mathbf{}\mathbf{of}\mathbf{}\mathbf{kinetic}\mathbf{}\mathbf{energy}\mathbf{,} \\ \mathbf{}\mathbf{and}\mathbf{}\mathbf{then}\mathbf{,}\mathbf{}\mathbf{using}\mathbf{}\mathbf{the}\mathbf{}\mathbf{given}\mathbf{}\mathbf{values}\mathbf{,} \\ \mathbf{}\mathbf{we}\mathbf{}\mathbf{can}\mathbf{}\mathbf{find}\mathbf{}\mathbf{the}\mathbf{}\mathbf{kinetic}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{particle}\mathbf{.} \\ \mathrm{Formula}: \\ \mathrm{Centripetal}\mathrm{force},\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \mathrm{Kinetic}\mathrm{energy},\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2 \end{gathered}$$

Explanation

$$\begin{gathered} Sincecentripetalforceisequaltomagneticforce,theexpressioncanbewrittenas: \\ F=\frac{m{v}^2}{r} \\ Fr=m{v}^2 \\ But,wehave, \\ KE=\frac{m{v}^2}{2} \\ Therefore, \\ KE=\frac{m{v}^2}{2} \\ =\frac{Fr}{2} \\ =\frac{\left(1.60\times {10}^{-17}\right)\left(26.1\times {10}^{-6}\right)}{2} \\ =2.09\times {10}^{-22}J. \\ Therefore,thekineticenergyoftheparticleis2.09\times {10}^{-22}. \end{gathered}$$