An electron (mass 9 × 10^–31 kg and charge 1.6 ×10^–19 C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9 × 10–4 T perpendicular to its direction of motion. We wish to apply an uniform electric E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10⁸ ms^–1)

To find the uniform electric field $\overrightarrow{E}$ required to prevent the electron from deflecting, we analyze the forces acting on the electron.

When a charged particle moves through both a magnetic field $\overrightarrow{B}$ and an electric field $\overrightarrow{E}$, the net electromagnetic force (Lorentz force) acting on it is given by:
$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

For the electron to pass undeflected, the net force must be zero ($\overrightarrow{F}=0$). This implies:
$q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})=0$
$\overrightarrow{E}=-(\overrightarrow{v}\times \overrightarrow{B})$

Since the electric field $\overrightarrow{E}$ is equal to $-(\overrightarrow{v}\times \overrightarrow{B})$, it must be perpendicular to both the velocity vector $\overrightarrow{v}$ and the magnetic field vector $\overrightarrow{B}$. Therefore, $\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$.

Because the velocity vector $\overrightarrow{v}$ is perpendicular to the magnetic field vector $\overrightarrow{B}$, the angle between them is 90 degrees. Thus, the magnitude of the electric field is:
$E=vB$

Given values:
Speed of light, $c=3\times {10}^8{\text{ms}}^{-1}$
Speed of the electron, $v=\frac{c}{100}=\frac{3\times {10}^8}{100}=3\times {10}^6{\text{ms}}^{-1}$
Magnetic field magnitude, $B=9\times {10}^{-4}\text{T}$

Substituting these values into the formula to find the magnitude of the electric field:
$E=(3\times {10}^6)\times (9\times {10}^{-4})$
$E=27\times {10}^2{\text{V m}}^{-1}$

Therefore, the electric field $\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$ and its magnitude is $27\times {10}^2{\text{V m}}^{-1}$.