In Thomson's experiment, the objective was to balance the forces acting on the electrons so that they experience no deflection. This means that the electric force and the magnetic force must cancel each other out.

Step-by-step explanation:

1. Forces Acting on the Electron:

   • The force due to the electric field is: $F_E = e \cdot E$ where $e$ is the charge of the electron and $E$ is the electric field magnitude.

   • The force due to the magnetic field is: $F_H = e \cdot v \cdot H$ where $v$ is the speed of the electron and $H$ is the magnetic field magnitude.

2. Condition for Zero Deflection: For zero deflection, the forces must balance, so: $F_E = F_H$ Substituting for the forces: $e \cdot E = e \cdot v \cdot H$ Canceling $e$ from both sides (since it's non-zero): $E = v \cdot H$ Solving for the speed of the electron $v$: $v = \frac{E}{H}$ This gives the speed in terms of the electric field and magnetic field.

3. Including the Constant $\mu_0$: The expression we want to prove includes the vacuum permeability $\mu_0$. The permeability of free space $\mu_0$ relates the electric and magnetic fields to the force on a charged particle. In Gaussian units, the relation is more straightforward, but in SI units, the factor $\mu_0$ comes into play.

   Therefore, the speed of the electron becomes: $v = \frac{1}{\mu_0} \cdot \frac{E}{H}$ where $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m}/\text{A}$ is the permeability of free space.

4. Numerical Calculation:

   Given:

   • $E = 7 \, \text{V/cm} = 700 \, \text{V/m}$ (convert to SI units),
   • $H = 5 \, \text{A/cm} = 500 \, \text{A/m}$ (convert to SI units),
   • $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m}/\text{A}$.

   Substituting into the equation: $v = \frac{1}{\mu_0} \cdot \frac{700}{500}$

Let me calculate that for you.The speed of the electrons, based on the given fields, is approximately $v = 1.11 \times 10^6 \, \text{m/s}$.

Explanation of Units:

• $E$ was given as 7 V/cm, which converts to 700 V/m.
• $H$ was given as 5 A/cm, which converts to 500 A/m.
• $\mu_0$ is the vacuum permeability, equal to $4\pi \times 10^{-7} \, \text{T}\cdot\text{m}/\text{A}$.
• The final speed is expressed in meters per second (m/s), which is the standard unit for velocity.

Visual Representation:

Let me describe the situation with the electrons, fields, and forces:

• A beam of electrons moves along a straight line.
• A uniform electric field $E$ is directed perpendicular to the beam.
• A uniform magnetic field $H$ is orthogonal to both the electron beam and the electric field.
• The electric field is adjusted so that the deflection due to both forces is canceled out, leading to no net deflection.

Would you like me to generate a picture of the setup and the forces involved?

Relative Questions:

1. How would the speed of the electron change if the magnetic field was doubled?
2. What would happen if the electric field was increased but the magnetic field remained the same?
3. How does the permeability of free space $\mu_0$ affect the electron's speed?
4. What would happen if the magnetic field and electric field were not perpendicular?
5. How is the speed of light related to $\mu_0$ and $\varepsilon_0$ (permittivity of free space)?

Tip:

Remember that in Thomson’s experiment, the key to determining the electron’s speed lies in balancing the electric and magnetic forces to achieve zero deflection.