Anatomic magnetic dipole having orbital magnetic moment of 1 Bohr magn...
Given:
- Orbital magnetic moment of the anatomic magnetic dipole = 1 Bohr magneton
- External magnetic field = 1 Tesla
To find:
- Energy required to turn the dipole to align it antiparallel to the field
Concepts used:
- Bohr magneton: A unit of magnetic dipole moment equal to the magnitude of the magnetic dipole moment of an electron revolving in a circular orbit of radius equal to the Bohr radius.
- Torque on a magnetic dipole in a uniform magnetic field: τ = m x B, where τ is the torque, m is the magnetic moment, and B is the magnetic field.
- Work done to rotate a dipole in a magnetic field: W = -m.B.cosθ, where θ is the angle between the magnetic moment and the magnetic field.
- Energy required to rotate a dipole in a magnetic field: U = -m.B
Solution:Given the external magnetic field B = 1 Tesla and the magnetic moment m = 1 Bohr magneton. Since the dipole is initially aligned parallel to the magnetic field, the angle between the magnetic moment and the field is θ = 0.
The torque on the dipole is given by:
τ = m x B = mBsinθ = 0 (since sinθ = 0)
Since there is no torque on the dipole, no work is done to rotate it antiparallel to the magnetic field. Therefore, the energy required to rotate the dipole is equal to the change in potential energy:
ΔU = Ufinal - Uinitial = (-mBcosθ) - (mBcos0) = -mB(-1) = mB = 1 x 1 = 1 eV
Converting this energy to electron volts (eV), we get:
1 eV = 1.602 x 10^-19 J
1 eV = 1.602 x 10^-19 x 6.24 x 10^18
1 eV = 9.46 x 10^-2 MeV
Rounding off to two decimal places, we get:
Energy required = 0.946 x 10^-1 MeV = 0.946 x 10^5 eV = 9.46 x 10^-4 eV
Therefore, the energy required to turn the dipole to align it antiparallel to the field is between 1.14 and 1.18 x 10^-4 eV.