A point charges Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity `omega`. This can be considered as equivalent to a loop carrying a steady current `(Q omega)/(2 pi)`. S 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 hte orbit remains constant. The application of hte 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 si known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a porportionality constant `lambda`.
The charge in the magnetic dipole moment associated with the orbit. at the end of the time interval of hte magnetic field charge, is

Options

A. `-gammaBQR^(2)`
B. `-gamma(BQR^(2))/2`
C. `gamma(BQR^(2))/2`
D. `gammaBQR^(2)`

Answer

B

Explanation

Megnetic dipole moment `M=gamma J`
`DeltaM=gammaDeltaJ`…(i)
`(DeltaJ)/(Deltat)=-Q(dB)/(dt).R/2R. rArrDeltaJ=-(QB)/2R^(2) rArr "so" DeltaM=(gammaQBR^(2))/2`
Alternate
`M/L=Q/(2m)`
`M=(Qomega)/(2pi)piR^(2)=(QomegaR^(2))/2`.
induced electric field is opposite to the `omega` so the charge is retarded.
`omega'=omega-alphat`
`omega'=omega-(QB)/2 1 , (a_(1)=QE//m),(alpha=(QE)/(mR)=Q/Rxx(BR)/(2m)=(QB)/(2m))`
`M_(f)=(Qomega'R^(2))/2=Q(omega-(QB)/(2m))R^(2)/2`
`Deltam=M_(f)-M_(i)=(QomegaR^(2))/2-(Q^(2)BR^(2))/(4m)-(QomegaR^(2))/(2)=-gamma(BQR^(2))/2`
Alternate
`M/L=Q/(2m)`
`M=(Qomega)/(2pi)piR^(2)=(QomegaR^(2))/2`.
`omega'=omega-alphat`
`omega'=omega-(QB)/2 1 , (a_(1)=QE//m),(alpha=(QE)/(mR)=Q/Rxx(BR)/(2m)=(QB)/(2m))`
`M_(f)=(Qomega'R^(2))/2=Q(omega-(QB)/(2m))R^(2)/2`
`Deltam=M_(f)-M_(i)=(QomegaR^(2))/2-(Q^(2)BR^(2))/(4m)-(QomegaR^(2))/(2)=-gamma(BQR^(2))/2`