Question

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 $\overrightarrow{B}$ of magnitude 9 × 10⁻⁴ T perpendicular to its direction of motion. We wish to apply a uniform electric $\overrightarrow{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10⁸ ms⁻¹)

• A. $\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$ and its magnitude is 27 × 10⁴ V m⁻¹
• B. $\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$ and its magnitude is 27 × 10² V m⁻¹
• C. $\overrightarrow{E}$ is parallel to $\overrightarrow{B}$ and its magnitude is 27 × 10² V m⁻¹
• D. $\overrightarrow{E}$ is parallel to $B$ and its magnitude is 27 × 10⁴ V m⁻¹

Answer 1

Answer: (B)

$\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$ and its magnitude is 27 × 10² V m⁻¹

Answer 2

The electron moves undeflected in the combined electric and magnetic fields. The net force on the electron must be zero. The forces acting on the electron are the electric force $\overrightarrow{F}_E = q\overrightarrow{E}$ and the magnetic force $\overrightarrow{F}_B = q(\overrightarrow{v} \times \overrightarrow{B})$.

For undeflected motion, the net force is zero: $\overrightarrow{F}_E + \overrightarrow{F}_B = 0$ $q\overrightarrow{E} + q(\overrightarrow{v} \times \overrightarrow{B}) = 0$ Since $q \neq 0$, we can divide by $q$: $\overrightarrow{E} + (\overrightarrow{v} \times \overrightarrow{B}) = 0$ $\overrightarrow{E} = -(\overrightarrow{v} \times \overrightarrow{B})$

This equation gives the required electric field $\overrightarrow{E}$. From this equation, we can see that $\overrightarrow{E}$ must be perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$, because $\overrightarrow{v} \times \overrightarrow{B}$ is a vector perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$. The magnitude of $\overrightarrow{E}$ is $E = |\overrightarrow{v} \times \overrightarrow{B}|$.

We are given that the magnetic field $\overrightarrow{B}$ is perpendicular to the direction of motion $\overrightarrow{v}$. So, the angle between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $90^\circ$. The magnitude of the cross product is $|\overrightarrow{v} \times \overrightarrow{B}| = vB \sin\theta$, where $\theta$ is the angle between $\overrightarrow{v}$ and $\overrightarrow{B}$. Since $\theta = 90^\circ$, $\sin\theta = 1$. So, the magnitude of $\overrightarrow{E}$ is $E = vB$.

The speed of the electron is given as $v = c/100$, where $c = 3 \times 10^8$ m/s. $v = \frac{3 \times 10^8 \text{ m/s}}{100} = 3 \times 10^6 \text{ m/s}$. The magnitude of the magnetic field is given as $B = 9 \times 10^{-4}$ T.

The magnitude of the electric field is: $E = vB = (3 \times 10^6 \text{ m/s}) \times (9 \times 10^{-4} \text{ T})$ $E = (3 \times 9) \times (10^6 \times 10^{-4}) \text{ V/m}$ $E = 27 \times 10^{6-4} \text{ V/m}$ $E = 27 \times 10^2 \text{ V/m}$

The direction of $\overrightarrow{E}$ must be perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$. Since $\overrightarrow{v}$ and $\overrightarrow{B}$ are perpendicular to each other, $\overrightarrow{v} \times \overrightarrow{B}$ is perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$. Thus, $\overrightarrow{E} = -(\overrightarrow{v} \times \overrightarrow{B})$ is also perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$.

Therefore, $\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$.