Calculating change in Magnetic Dipole Moment

AI Thread Summary
The discussion focuses on calculating the change in magnetic dipole moment (Δm) when a charge moves faster in a circular path. The magnetic dipole moment is defined as m=IA, where I is the current and A is the area. The relationship between current and speed is established, leading to the formula Δm=1/2 qΔvr. Questions arise regarding the application of this formula to single atoms versus solids, with clarification that magnetic moments are relevant for describing torque in magnetic fields. The conversation emphasizes the importance of understanding magnetic moments in the context of both current-carrying loops and permanent magnets.
iScience
Messages
466
Reaction score
5
In one moment in time you have a charge going about in a circle of radius r. in another moment in time the charge goes in the same circle but faster, where Δv is the difference in speed. i want to find the change in the magnetic dipole moment Δm.if magnetic dipole moment m=IA (assuming uniform I), how do you go from

$$m=IA ...to... Δm=\frac{1}{2}qΔvr ?$$Thanks in advance
 
Physics news on Phys.org
well, what's the current as a function of speed v?
 
I(v)=\lambdav
 
iScience said:
I(v)=\lambdav

The problem says "a charge" so I would assume a single charge q rather than a charge distribution λ, so say the charge goes around N times/s, which is 2πRN m in 1 s, which says velocity v = 2πRN, which says the current is how much?
 
oh... got it now! thanks!

but another question: i can't say this for a single atom since current exists only on the outside, but for the case of a solid, wouldn't "IA" be more of a flux than a moment? and even for the case of a single atom, i thought moments were supposed to be some quantity times a distance, not an area.
 
iScience said:
oh... got it now! thanks!

but another question: i can't say this for a single atom since current exists only on the outside, but for the case of a solid, wouldn't "IA" be more of a flux than a moment? and even for the case of a single atom, i thought moments were supposed to be some quantity times a distance, not an area.

Not sure I know what you mean by invoking atoms and solids, but the main point of the magnetic moment μ = IA is that, in a B field, there is a moment (torque) τ applied to the coil or whatever medium carrying the current such that τ = μ x B.

You can easily show the validity of this formula by considering the torques on a rectangle of wire, sides 2a and 2b, carrying current I in a field B parallel to any side. For other shapes, integration would be necessary.

Magnetic moments are especially handy in describing permanent magnets. A p.m. with mag. moment μ will see a torque as per the above. So μ can be measured by pointng the p.m. at right angles to the B field and recording the torque. The A vector points from the S to the N pole. I offhand don't know another way to quantify permanent magnets.


Take a look at http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html

and

http://en.wikipedia.org/wiki/Magnetic_moment
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top