# Calculating the rate of change of Magnetic Field strength [closed]

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?

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 is simply

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

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

## closed as off-topic by user10851, Gert, John Rennie, user36790, CuriousOneMay 16 '16 at 0:11

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• $d r^{-1}/ dt = -r^{-2} dr/dt = r^{-2}v$. – Rob Jeffries May 14 '16 at 9:01
• should the negative sign out the front have disappeared at the end there? – Lincoln77 May 14 '16 at 9:04
• It depends exactly how v is defined (speed or velocity?). $dr/dt$ is negative right? Do you think $dB/dt$ is positive or negative? – Rob Jeffries May 14 '16 at 9:08
• I think the change in B is positive, since the wire is getting closer to the loop? – Lincoln77 May 14 '16 at 9:10
• So, if I define $v$ to be the closing speed, I think you'll find the sign is correct... – Rob Jeffries May 14 '16 at 9:13

Look at time $t=0$, magnetic field at a particular point X, which is a distance $r$ from the wire is $B(0) = \frac{\mu_0 I}{2 \pi r}$. At a small time $\Delta t$ the wire has moved a distance $v \Delta t$, so that this point is now only a distance $r - v \Delta t$ away from the wire, and the value of the magnetic field changes because you are nearer) $B(\Delta t) = \frac{\mu_0 I}{2 \pi (r - v \Delta t)}$. Then assume that $\Delta t$ is small, and expand the expression $\frac{1}{1+x} \approx 1 - x$ for the relevant $x$, so that you can find $\Delta B = B(\Delta t) - B(0)$.
However, this is non-relativistic and the "best" way (in the sense more accurate, but it depends on the level of electromagnetism you are at) to do it is to determine the magnetic field at the point X in the rest frame of the wire (giving $B = \frac{\mu_0 I}{2 \pi r}$) and then boost to a frame moving with speed $v$ wrt the wire.
• Hi, thanks for your answer, I followed you up to the point where you said $\frac{1}{1+x} \approx 1 - x$, can you explain to me where this is coming from, i dont understand it? – Lincoln77 May 14 '16 at 8:17
• $\frac{1}{r - v \Delta t}= \frac{1}{r} \times \frac{1}{1 - v \Delta t/r}$, so you can identify $x = - v \Delta t/r$. – jim May 14 '16 at 8:19
• All I have done is calculate the value of the magnetic field at a fixed point, X. At time $t=0$ the distance of the fixed point is a distance $r$ away from the wire, then at a small time $\Delta t$ later the wire has moved a distance $v \Delta t$ so that the point X is now only a distance $r - v \Delta t$ away from the wire. – jim May 14 '16 at 8:31
• hmm, ok, i can see what youve done, I suppose I should have put it in my question but, one of the hints we are given is that $v=\frac{dx}{dt}$ and so, while i think you are right, I'm not sure if the online quiz thing will accept the answer given by $\Delta B = B(\Delta t) - B(0)$, and I think we are supposed to derive an equation for the change in B using related rates. Thats what I'm unsure how to do. In any case thank you for your help. – Lincoln77 May 14 '16 at 8:47
$$\frac{d|B|}{dt}=\frac{d|B|}{dr}\frac{dr}{dt} = -\frac{μ_{0}I}{2πr^2}v$$