Question

An electron is placed in a magnetic field $\overrightarrow{B}$ that is directed along a$z$axis. The energy difference between parallel and antiparallel alignments of the $z$component of the electron’s spin magnetic moment with$\overrightarrow{B}$is$6.00\times {10}^{-25}$J. What is the magnitude of$\overrightarrow{B}$?

Short answer

The magnitude of the magnetic field is$32.36\text{mT}$

Step-by-step solution

Identification of given data

Theenergy difference between parallel and antiparallel alignments of the $z$component of the electron’s spin magnetic moment is, $\Delta U=6.0\times {10}^{-25}\text{J}$

Definition of Potential Energy of Spin Magnetic Moment and expression for the magnitude of the magnetic field

The potential energy of the spin magnetic dipole moment kept in an external magnetic field is given by the dot product of the spin magnetic dipole moment and the external magnetic field.The expression for the magnitude of the magnetic field is given as follows,

$B=\frac{\Delta U}{2{\mu }_B}$

Here, $\Delta U$is theenergy difference between parallel and antiparallel alignments, and ${\mu }_B$is themagnetic dipole moment with the value$9.27\times {10}^{-24}\mathrm{J}/\mathrm{T}$.

Determination of the Magnitude of the Magnetic Field

Substitute all the values in the expression for the magnitude of the magnetic field.

$$\begin{gathered} B=\frac{6.0\times {10}^{-25}\text{J}}{2\times 9.27\times {10}^{-24}\text{J/T}} \\ =\frac{3\times {10}^{-1}}{9.27}\text{T} \\ =0.3236\times {10}^{-1}\text{T}\times \frac{1000\text{mT}}{1\text{T}} \\ =32.36\text{mT} \end{gathered}$$

Thus, the magnitude of the magnetic field is$32.36\text{mT}$ .