How can I derive a formula for change in B field strength?

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SUMMARY

The discussion revolves around deriving the formula for the rate of change of magnetic field strength at the center of a circular wire loop due to a long wire carrying current I, moving towards the loop with speed v. The magnetic field strength at a distance from the wire is given by the equation |B| = (μ₀I)/(2πr). The correct formula for the rate of change of the magnetic field strength is derived as d|B|/dt = (μ₀IV)/(2πr²) after applying implicit differentiation and recognizing the relationship between dr/dt and the velocity of the wire. The initial incorrect assumption was clarified through dimensional analysis and proper differentiation.

PREREQUISITES
  • Understanding of Ampere's Law and its application in magnetic fields
  • Familiarity with implicit differentiation in calculus
  • Knowledge of magnetic field equations, specifically |B| = (μ₀I)/(2πr)
  • Basic concepts of electromagnetism, including current and magnetic field interactions
NEXT STEPS
  • Study the implications of Ampere's Law in different geometries
  • Learn about implicit differentiation techniques in calculus
  • Explore the relationship between current, magnetic fields, and motion in electromagnetism
  • Investigate applications of magnetic field changes in real-world scenarios, such as inductance
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Students studying electromagnetism, physics educators, and anyone interested in understanding the dynamics of magnetic fields in relation to moving currents.

RobertColman
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Member advised to use the homework template for posts in the homework sections of PF.
I'm a bit stuck on this question (which is homework so hints are more welcome than outright answers). The question is:

A very long wire carrying a current I is moving with speed v towards a small circular wire loop of radius r. The long wire is in the plane of the loop and is too long to be entirely shown in the diagram.

lQAjR.png


The strength of a magnetic field a distance x from a long wire is
$$|B|=\frac{μ_{0}I}{2πr}$$
What is the equation for the rate of change of the strength of the magnetic field at the centre of the loop?

As a hint we are given that $$V = \frac{dx}{dt}$$

Now, I can see that the equation for the strength of the field comes from amperes law, and is essentially the magnetic field along a loop around the wire, divided by the circumference of that loop.

So it makes sense to me that given everything else in the equation is constant, delta B should come straight from the change in the circumference of a circle as the radius shrinks. Now since

$$\frac{dC}{dt}=2\pi\frac{dr}{dt}$$

and since in this case $\displaystyle{\frac{dr}{dt}}$ is simply the velocity of the wire, it seems to me that the change in the field strength at the center of the loop should simply be

$$\frac{d|B|}{dt}=\frac{μ_{0}I}{2πV}$$

however I'm getting that this answer is wrong. Can anyone explain where I've made a mistake?

Thanks for your help.
 
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RobertColman said:
however I'm getting that this answer is wrong.
Just from the dimensional analysis, one can tell that it is incorrect.
The rate of change of the strength of the B field at a fixed point is
$$
\frac{d}{dt}|B|
$$
Now, plug in the equation for ##|B|## into the time derivative.
 
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So, would you expect to get $$\frac{μ_{0}Ir^{2}-2\pi V}{2\pi r^{2}}$$ ?
 
RobertColman said:
So, would you expect to get $$\frac{μ_{0}Ir^{2}-2\pi V}{2\pi r^{2}}$$ ?
Is that what you get after differentiating ##|B|## with respect to ##t##?
 
That's what 'I' got, whether or not it is correct is what I was asking? should it be a plus instead of a minus?

as in, I used implicit differentiation to turn 1/r into -1/r^2 * dr/dt. since in this case dr/dt = -V that gives V/r^2 etc etc
 
ok i think i got it, i forgot that the constants I pulled out the front are being multiplied back in, so it should be $$\frac{μ_{0}IV}{2\pi r^{2}}$$?
 
RobertColman said:
ok i think i got it, i forgot that the constants I pulled out the front are being multiplied back in, so it should be $$\frac{μ_{0}IV}{2\pi r^{2}}$$?
Yes.
 
ok, Thank you so much.
I am way too stressed right now, and you were a big help :)
 

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