An electron orbits a proton at a radius of 5.38 10-11 m. What is the magnitude of the magnetic field at the proton due to the orbital motion of the electron?
To find the magnitude of the magnetic field at the proton due to the orbital motion of the electron, you can use the formula for the magnetic field of a moving charge in a circular orbit. Here's how you can calculate it:
1. Start with the formula for the magnetic field of a moving charge in a circular orbit:
B = μ₀ * (q * v) / (2 * π * r)
Where:
- B is the magnetic field
- μ₀ is the permeability of free space (μ₀ = 4π × 10⁻⁷ T·m/A)
- q is the charge of the moving particle
- v is the velocity of the moving particle
- r is the radius of the circular orbit
2. Determine the values for the variables in the formula:
- Charge of the electron (q) = -1.6 × 10⁻¹⁹ C (since the proton has opposite charge)
- Velocity of the electron (v) = speed of light (since electrons in atomic orbitals move at speeds close to the speed of light, use v = 3 × 10⁸ m/s for simplicity)
- Radius of the orbit (r) = 5.38 × 10⁻¹¹ m (given in the question)
3. Substitute the values into the formula:
B = (4π × 10⁻⁷ T·m/A) * (-1.6 × 10⁻¹⁹ C * 3 × 10⁸ m/s) / (2 * π * 5.38 × 10⁻¹¹ m)
4. Calculate the magnetic field:
B ≈ -2.287 × 10⁻⁵ T
So, the magnitude of the magnetic field at the proton due to the orbital motion of the electron is approximately 2.287 × 10⁻⁵ T.
Oliver Heaviside, in 1888, first solved this question for moving charges.
B=mu*q/4PI * v/r^2
http://bama.ua.edu/~stjones/puzzles.htm