Paragraph: A point charge Q is moving in a circular orbit of radius...
Paragraph:
A point charge Q is moving in a circular orbit of radius R in the x−y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current Qω2π. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ.
Question:
The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is
A point charge Q is moving in a circular orbit of radius R in the x−y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current Qω2π. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ.
Question:
The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is
Solution:
Magnetic dipole moment M=γJ
ΔM=γΔJ ...(1)
ΔJΔt=-QdBdt·R2R
ΔJ=-QB2R2
So ΔM=-γQBR22
Alternate

ML=Q2m
M=Qω2ππR2=QωR22
induced electric field is opposite to the ω so the charge is retarded.
ω′=ω-αt
ω′=ω-QB21(at=QE/m),(α=QEmR=QR×BR2m=QB2m)
Mf=Qω′R22=Q(ω-QB2m)R22
∆M=Mf-Mi=QωR22-Q2BR24m-QωR22=-γBQR22
ΔM=γΔJ ...(1)
ΔJΔt=-QdBdt·R2R
ΔJ=-QB2R2
So ΔM=-γQBR22
Alternate

ML=Q2m
M=Qω2ππR2=QωR22
induced electric field is opposite to the ω so the charge is retarded.
ω′=ω-αt
ω′=ω-QB21(at=QE/m),(α=QEmR=QR×BR2m=QB2m)
Mf=Qω′R22=Q(ω-QB2m)R22
∆M=Mf-Mi=QωR22-Q2BR24m-QωR22=-γBQR22
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