Question

In Fig 28-32, an electron accelerated from rest through potential difference ${\mathbf{V}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{kV}}$enters the gap between two parallel plates having separation $\mathbf{d}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{mm}}$and potential difference ${\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{100}\mathbf{ }\mathbf{\text{V}}$. The lower plate is at the lower potential. Neglect fringing and assume that the electron’s velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap?

Short answer

The potential difference is, .The uniform magnetic field is $\overrightarrow{B}=-\left(2.67\times {10}^{-4} \text{T}\right)\hat{k}$

Step-by-step solution

Step 1: Identification of the given data

The potential difference is,$V_1=1.0 \text{kV}$.

The potential difference is,$V_2=100 \text{V}$.

The distance between plates is, $d=\text{20} \text{mm}$.

Determining the concept

The net force acting on the electron must be zero if it is to be in straight line motion between the parallel plates. Also use law of conservation of energy,

The law of conservation of energy states that the amount of energy is neither created nor destroyed.

Formula are as follow:

$E=\frac{V}{d}$

$F_e=qE$

$F_m=qvB$

Where, F is force, v is velocity, E is electric field, B is magnetic field, q is charge of particle, d is distance between plates.

Determining the uniform magnetic field

To find the speed of the electron:

Applying law of conservation of energy,

$$\begin{gathered} KE=PE \\ \frac{1}{2}m{v}^2=q{V}_1 \\ v=\sqrt{\frac{2q{V}_1}{m_e}} \end{gathered}$$

To find electric field,

$E=\frac{V_2}{d}$

To find magnetic field,

Here, magnetic force on the electron balances the electric force on the electron.

$$\begin{gathered} F_e=F_m \\ qE=qvB \\ B=\frac{E}{v} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} B=\frac{\frac{V_2}{d}}{\sqrt{\frac{2q{V}_1}{m_e}}} \\ B=\frac{\frac{100 \text{V}}{20\times {10}^{-3} \text{m}}}{\sqrt{\frac{2\times \left(1.0\times {10}^3 \text{V}\right)\times 1.6\times {10}^{-19} \text{C}}{9.1\times {10}^{-31} \text{kg}}}} \end{gathered}$$

$B=2.67\times {10}^{-4} \text{T}$

To make the electron move with constant velocity, the force due to electric field of the electron must be balanced by the force due to magnetic field on the electron. Since the force due to electric field is along $-\hat{j}$, the magnetic field should be along $-\hat{k}$to counter the electric force.

$B=-\left(2.67\times {10}^{-4} \text{T}\right)\hat{k}$

Hence, the uniform magnetic field that allows the electron to travel in a straight line in the gap is $B=-\left(2.67\times {10}^{-4} \text{T}\right)\hat{k}$.

Therefore, by using the formula of electric and magnetic field and the law of conservation of energy, the uniform magnetic field can be determined.