Question

An electron has a spin magnetic moment of magnitude \(\mu=9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2} .\) Consequently, it has energy associated with its orientation in a magnetic field. If the difference between the energy of an electron that is "spin up" in a magnetic field of magnitude \(B\) and the energy of one that is "spin down \(^{\infty}\) in the same magnetic field (where "up" and "down" refer to the direction of the magnetic field) is \(9.460 \cdot 10^{-25} \mathrm{~J},\) what is the field magnitude, \(B ?\)

Short answer

Answer: The approximate magnitude of the magnetic field is \(B \approx 1.018 \mathrm{~T}\).

Step-by-step solution

Write down the given information on energy difference and spin magnetic moment

We are given the energy difference \({\Delta}E = 9.460 \cdot 10^{-25} \mathrm{~J}\) and spin magnetic moment \(\mu=9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\).

Write down the formula for energy difference

The energy difference formula between "spin up" and "spin down" states is given by \({\Delta}E = \mu B\).

Solve for the magnetic field magnitude B

Using the given values and energy difference formula, we can now solve for B: \(9.460 \cdot 10^{-25} \mathrm{~J} = (9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}) B\) Now, divide both sides of the equation by \(9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\): \(B = \frac{9.460 \cdot 10^{-25} \mathrm{~J}}{9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}}\)

Calculate the magnetic field magnitude B

Now, we can calculate the value of B: \(B = \frac{9.460 \cdot 10^{-25}}{9.285 \cdot 10^{-24}} ≈ 1.018 \mathrm{~T}\) Therefore, the magnetic field magnitude is approximately \(B \approx 1.018 \mathrm{~T}\).