Answer to Question #88618 in Electricity and Magnetism for Sridhar

Question

a beam of electrons enters in uniform magnetic field of flux density 1.5 Tesla the energy difference between the electrons whose spin are parallel and antiparallel to the field is
1)1.74×10^-4 eV
2)0.87×10^-4 ev
3)3.48×10^-4 ev
4)2.5×10^-4 ev

Answer 1

The potential energy of a particle possessing a magnetic (dipole) moment and interacting with a uniform magnetic field can be obtained as follows:

$$U = - \vec{\mu} \cdot \vec{B},$$

where \mu is the magnetic moment of a particle which is parallel to spin. Hence, the difference between the potential energies of particles whose spin is parallel and antiparallel to the field is

$$\Delta U = 2 \mu B$$

 In case of an electron

$$\mu \approx \mu_B \approx 5.79 \cdot 10^{-5} \, \frac{eV}{T},$$

where \mu_B is the Bohr magneton.

Substituting the numerical values, we obtain:

$$\Delta U = 2 \cdot 5.79 \cdot 10^{-5} \cdot 1.5 \approx 1.74 \cdot 10^{-4} \, eV$$

Answer: 1) 1.74×10^-4 eV.