Question

A particle with mass$\mathbf{1}\mathbf{.}\mathbf{81}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{k}\mathit{g}$and a charge of has$\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathit{C}$, at a given instant, a velocity$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{V}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}$.What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field$\mathit{B}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{63}\mathbf{T}\mathbf{)}\mathit{i}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{980}\mathbf{T}\mathbf{)}\hat{\mathbf{j}}$?

Short answer

The acceleration is$\stackrel{\rightharpoonup }{a}=-(0.330m/s^2)\hat{k}$

Step-by-step solution

Important Concepts

The magnetic field force is given by

$$\begin{gathered} {\mathit{F}}_{\mathbf{B}}\mathbf{=}\mathit{q}\mathbf{(}\mathit{v}\mathbf{\times }\mathit{b}\mathbf{)} \\ \mathit{F}\mathbf{=}\mathit{q}\mathit{v}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta } \end{gathered}$$

Where q is the charge of the particle, V is the velocity and B is the magnetic field

Determination of magnitude and direction of the particles

By the second law of motion we know that the force on a body is

$\stackrel{\rightharpoonup }{F}=m\stackrel{\rightharpoonup }{a}$

And the magnetic field force is given by

$F=q(v\times B)$

Equating the two we get

$$\begin{gathered} m\stackrel{\rightharpoonup }{a}=q\left(v\times B\right) \\ \stackrel{\rightharpoonup }{a}=\frac{q\left(v\times B\right)}{m} \\ \mathrm{Substitute}\mathrm{all}\mathrm{the}\mathrm{value}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation} \\ \stackrel{\rightharpoonup }{\mathrm{a}}=\frac{\left(1.22\times {10}^{-8}\mathrm{C}\right)\left(\left(3.0\times {10}^4\mathrm{m}/\mathrm{sj}\right)\times \left(1.63\mathrm{t}\stackrel{\rightharpoonup }{\mathrm{i}}+0.980\mathrm{T}\stackrel{\rightharpoonup }{\mathrm{j}}\right)\right)}{1.81\times {10}^{-3}\mathrm{kg}} \\ \stackrel{\rightharpoonup }{\mathrm{a}}=-(0.330\mathrm{m}/{\mathrm{s}}^3)\hat{\mathrm{k}} \\ \mathrm{Hence},\mathrm{the}\mathrm{acceleration}\mathrm{is}\stackrel{\rightharpoonup }{\mathrm{a}}=-(0.330\mathrm{m}/{\mathrm{s}}^3)\hat{\mathrm{k}}. \end{gathered}$$