Question

Question: An electric field of$\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{kV}\mathbf{/}\mathbf{m}$and a perpendicular magnetic field of $\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{}\mathbf{T}$act on a moving electron to produce no net force. What is the electron’s speed?

Short answer

The electron’s speed is $\left|v\right|=3.75\mathrm{x}{10}^3\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given

$$\begin{gathered} \varphi =90.0^{\circ } \\ E=1.50\frac{\mathrm{kV}}{\mathrm{m}}=1.50\mathrm{x}{10}^3\mathrm{V}/\mathrm{m} \\ B=0.400\mathrm{T} \\ F=0 \end{gathered}$$

Determining the concept

Find the velocity of an electron by inserting the given values in the formula for electromagnetic force.

Formulae are as follow:

$F=eE+evB\sin \varphi$

Where, F is force, v is velocity, E is electric field, B is magnetic field, e is charge of particle.

Determining the speed of electron

The electromagnetic force experienced by an electron is,

$F=eE+evB\sin \varphi$

Inserting given values,

$$\begin{gathered} 0=\left(-1.6\times {10}^{-19}\right)\times \left(1.50\times {10}^3\right)+\left(-1.6\times {10}^{-19}\right)\times \mathrm{v}\times \left(0.400\times \sin {90}^{\circ }\right) \\ 0=\left(-1.6\times {10}^{-19}\right)\times \left(1.50\times {10}^3\right)-\left(1.6\times {10}^{-19}\right)\times \mathrm{v}\times \left(0.400\times \sin {90}^{\circ }\right) \\ \left|\mathrm{v}\right|=\frac{\left(1.50\times {10}^3\right)}{0.400} \\ \left|\mathrm{v}\right|=3.75\times {10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the electron’s speed is $\left|\mathrm{v}\right|=3.75\times {10}^3\mathrm{m}/\mathrm{s}$.

Therefore, the velocity of an electron can be determined by inserting the given values in the formula for electromagnetic force.