Question

An electron experiences a magnetic force of magnitude$\mathbf{4}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{N}$when moving at an angle of 60.0° with respect toa magnetic field of magnitude$\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{T}$. Find the speed ofthe electron.

Short answer

The velocity is$\stackrel{\rightharpoonup }{\mathrm{v}}=9.49\times {10}^6\mathrm{m}/\mathrm{s}$

Step-by-step solution

The significance of the magnetic field

The magnetic field force is given by
$$\begin{gathered} {\mathbf{F}}_{\mathbf{B}}\mathbf{=}\mathbf{q}\left(v\times B\right) \\ \mathbf{}\mathbf{F}\mathbf{=}\mathbf{qvBsin\theta } \end{gathered}$$

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

Determination of the speed of the electron

We know that the magnetic field force is given by

$$\begin{gathered} \mathrm{F}=\mathrm{q}(\mathrm{v}\times \mathrm{B}) \\ \mathrm{F}=\mathrm{q}\left|\mathrm{v}\right|\left|\mathrm{B}\right|\sin \theta \end{gathered}$$

Rearranging we get

$\left|V\right|\frac{F}{q\left|B\right|sin\theta }$

Plug the values

$$\begin{gathered} V=\frac{4.60\times {10}^{-15}N}{\left(1.6\times {10}^{-19}C\right)\left(3.50\times {10}^{-3}T\right)sin60} \\ v=9.{49}^6\mathrm{m}/\mathrm{s} \\ \mathrm{Hence},\mathrm{the}\mathrm{velocity}\mathrm{is}\stackrel{\rightharpoonup }{\mathrm{v}}=9.49\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$