Question

Write down the equation for Lorentz force. An electron passes undeviated from a place where an electric field \(5 \times 10^4~\text{V/m}\) and magnetic field of \(5 \times 10^{-2}~\text{weber/m}^2\) are applied. Calculate the velocity of the electron.

Hint:

For the electron to pass through undeviated, the electric and magnetic forces must be equal. Use the relation \(E = vB\) to find the velocity when both fields are applied.

Answer 1

The Lorentz force is the total force experienced by a charged particle due to both electric and magnetic fields. It is given by the following equation:
\[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \] Where:
- \(\vec{F}\) is the Lorentz force,
- \(q\) is the charge of the particle,
- \(\vec{E}\) is the electric field,
- \(\vec{B}\) is the magnetic field,
- \(\vec{v}\) is the velocity of the particle.
When an electron passes undeviated through the region of the electric and magnetic fields, it means that the electric force exactly cancels out the magnetic force. In such a situation, the electric force is equal to the magnetic force. The electric force \(F_{\text{electric}}\) is given by:
\[ F_{\text{electric}} = qE \] Where:
- \(E\) is the magnitude of the electric field.
The magnetic force \(F_{\text{magnetic}}\) is given by:
\[ F_{\text{magnetic}} = qvB \] Where:
- \(v\) is the velocity of the electron,
- \(B\) is the magnitude of the magnetic field.
For the electron to pass undeviated, the forces must balance each other out, i.e., the electric force equals the magnetic force:
\[ F_{\text{electric}} = F_{\text{magnetic}} \] Thus, we have:
\[ qE = qvB \] Canceling the charge \(q\) from both sides:
\[ E = vB \] Now, solving for the velocity \(v\):
\[ v = \frac{E}{B} \] Substitute the given values: - \(E = 5 \times 10^4~\text{V/m}\),
- \(B = 5 \times 10^{-2}~\text{weber/m}^2\),
We get:
\[ v = \frac{5 \times 10^4}{5 \times 10^{-2}} = 10^6~\text{m/s} \] Thus, the velocity of the electron is \(v = 1 \times 10^6~\text{m/s}\).
This result means that the electron will travel with a velocity of \(1 \times 10^6~\text{m/s}\) in order to pass through the fields without deviation.