Question

An electron with kinetic energy 2.5keVmoving along the positive direction of an xaxis enters a region in which a uniform electric field $\overrightarrow{\mathbf{B}}$of magnitude 10kV/mis in the negative direction of the yaxis. A uniform magnetic field is to be set up to keep the electron moving along the xaxis, and the direction of $\overrightarrow{\mathbf{B}}$ is to be chosen to minimize the required magnitude of $\overrightarrow{\mathbf{B}}$. In unit-vector notation, what$\overrightarrow{\mathbf{B}}$should be set up?

Short answer

The magnetic field $\overrightarrow{B}$is $(-3.4\times {10}^{-4}T)\hat{k}.$

Step-by-step solution

Identification of given data

1. The kinetic energy is K=2.5keV along the +x axis direction
2. The uniform electric field of magnitude is E=10kV/m in the -y axis direction
3. The uniform magnetic field to be set up to keep electron moving along x axis.
4. The magnitude of the uniform magnetic field is to be chosen minimum.

Understanding the procedure

We can find the equation for the magnitude of the magnetic field by equating electric force and magnetic force. By using the equation for kinetic energy, we can find the value of speed vand we can substitute this value in the equation of the magnetic field.By considering the direction of the magnetic field, electric field, and the velocity of electrons, we can find the direction of the magnetic field.

Formula:

1. The electric force,$F_e=eE$
2. The magnetic force,$F_B=qvBsin\theta$

The kinetic energy,$K=\frac{1}{2}{m}_e{v}^2$

Determining that what B→ should be set up in unit-vector notation

The electric force is given by, $F_e=eE$

Where E is the magnitude of the electric field.

The magnetic force is given by

localid="1663953446864" $F_B=qvBsin\theta$

By equating the magnitude of the electric force and the magnetic force we can obtain the magnitude of the magnetic field.

Therefore,$eE=qvB\sin \theta$

Since,q=e

$.:B=\frac{E}{v\sin \theta }$

As the magnitude of the magnetic field is to be chosen minimum.

The field is minimum when sin$\theta$is the largest i.e.$\theta ={90}^{\circ },\sin \theta =1$

Thus, the minimum magnetic field is

localid="1663953450454" $B=\frac{E}{v}$

Now the kinetic energy is given by

$K=\frac{1}{2}{m}_e{v}^2$

Therefore, the speed is

$$\begin{gathered} v=\sqrt{\frac{2K}{m_e}} \\ =\sqrt{\frac{2\times (2.5\times {10}^{3}eV)\times (1.60\times {10}^{-19}J/eV}{(9.11\times {10}^{-31}kg)}} \\ =2.96\times {10}^{7}m/s \end{gathered}$$

Thus, the magnetic field is given by

$$\begin{gathered} B=\frac{E}{v} \\ =\frac{10\times {10}^{3}V/m}{2.96\times {10}^{7}m/s} \\ =3.4\times {10}^{-4}T \end{gathered}$$

The direction of the magnetic field must be perpendicular to both the electric field $(-\hat{j})$and the velocity of the electron(+i^). Since the electric force $\left(F_e\right)⃗=-eE⃗$points in the (+j) direction, the magnetic force $\left(F_e\right)⃗=-e(v⃗\times B⃗)$points in the (-j) direction. Hence, the direction of the magnetic field is (-k).

In unit vector notation, the magnetic field is $B=3.4\times {10}^{-4}T$.