Question Details

An electron (mass $9 \times 10^{- 31} kg$ and charge $1.6 \times 10^{- 19} C$ ) moving with speed $\frac{c}{100}$ (c = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{- 4} T$ perpendicular to its direction of motion.

We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^{8} m s^{- 1}$ )

Options

$\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^{4} V m^{- 1}$

$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{4} V m^{- 1}$

$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{2} V m^{- 1}$

$\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^{2} V m^{- 1}$

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Correct Answer :

Option C

$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{2} V m^{- 1}$

E is perpendicular to B and its magnitude is 27×10^2 V m⁻¹

Solution :

To keep the electron moving in a straight line we need the total electromagnetic force on it to vanish:

$q\,( \mathbf{E} + \mathbf{v}\times\mathbf{B} ) = \mathbf{0}$

Since the charge of the electron is $q = -\,1.6\times10^{-19}\,\text{C}$ , the condition simplifies to

$\mathbf{E} = -\,\mathbf{v}\times\mathbf{B}$

The velocity is given as one hundredth of the speed of light:

$\displaystyle v = \frac{c}{100} = \frac{3\times10^{8}\,\text{m/s}}{100}=3\times10^{6}\,\text{m/s}$

The magnetic field magnitude is

$B = 9\times10^{-4}\,\text{T}$

Because the problem states that the magnetic field is perpendicular to the electron’s direction of motion, the angle between $\mathbf{v}$ and $\mathbf{B}$ is 90°, and therefore the magnitude of the cross product is simply the product of the magnitudes:

$\lvert \mathbf{v}\times\mathbf{B}\rvert = v\,B\sin 90^{\circ}=v\,B$

Insert the numerical values:

$\begin{aligned}
v\,B &= (3\times10^{6}\,\text{m/s})\;(9\times10^{-4}\,\text{T})\\
&= 27\times10^{2}\,\frac{\text{V}}{\text{m}}\\
&= 2.7\times10^{3}\,\frac{\text{V}}{\text{m}}.
\end{aligned}$

Thus the required electric field has magnitude

$\lvert\mathbf{E}\rvert = 27\times10^{2}\,\text{V\,m}^{-1}$

and its direction is opposite to $\mathbf{v}\times\mathbf{B}$ . Since $\mathbf{v}\times\mathbf{B}$ is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$ , the electric field is also perpendicular to the magnetic field.

Therefore, the correct statement is:

$\mathbf{E}\;\text{is perpendicular to}\;\mathbf{B}\;\text{and its magnitude is}\;27\times10^{2}\,\text{V\,m}^{-1}$