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The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²
<h3> Magnetic dipole moment of the bar magnet</h3>
The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;
where;
- B is magnetic field
- m is dipole moment
- μ is permeability of free space
m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)
m = 1.2 Am²
The complete question is below:
What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.
Learn more about dipole moment here: brainly.com/question/27590192
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Answer:
showm
Explanation:
Consider a dipole having magnetic moment 'm' is placed in magnetic field then the torque exerted by the field on the dipole is
Now to rotate the dipole in the field to its final position the work required to be done is
Minimum energy mB is for the case when m is anti parallel to B.
Minimum energy -mB is for the case when m is parallel to B.