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

• RobertColman
In summary, the equation for the rate of change of the strength of the magnetic field at the center of the loop is: $\frac{d}{dt}|B|=\frac{μ_{0}I}{2πV}$.

#### RobertColman

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. 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?

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.

• Anuran Pal
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 :)