Magnetic flux of sliding bar on rails

  • #1

Homework Statement



"A metal bar which runs on 2 long parallel rails are connected to a charged capacitor with capacitance C and a resistor with resistance R. Assume no friction and perfect conductors. The rails are cylindrical of radius R separated by distance d. The bar is a distance x along the rails where x >> d >> r.

What is the change in the magnetic flux through the center of the circuit if the current I is held constant but the bar is moving with a velocity v. For this part assume the rails are half infinite wires. Careful, it is not obvious that you can ignore the flux contribution from the sliding bar. You have to be clever here."

Homework Equations



My question is why exactly can we ignore the flux contribution from the bar?

It is seems intuitively obvious to me that we can since since x >> d. However my professor generally wants us to be quantitative on things like this and I am not exactly sure how to go about showing this quantitatively. Also I am not sure I am even going about the entire problem correctly.

The Attempt at a Solution



So this is what I have so far. The problem states that we assume the rails are half infinite wires.

The magnitude of the magnetic field of half infinite wire is [itex] |\vec{B}|= \frac{\mu_0 I}{4\pi s} [/itex] where [itex]s[/itex] is the perpendicular distance from the wire

So the flux is
[itex]
\begin{align*}
\Phi &= \int_{circuit} (\vec{B_{top}} + \vec{B_{bottom}}) \cdot d\vec{a} \\
&= \int_0^x \int_r^{d-r} \Biggr( \frac{\mu_0 I}{4\pi (y)} + \frac{\mu_0 I}{4\pi (d-y)} \Biggr) dx dy\\
&=\frac{\mu_0 I x}{2\pi} \ln \Biggr( \frac{d-r}{r} \Biggr)
\end{align*}
[/itex]

Now taking the time derivative

[itex]
\begin{align*}
\frac{d\Phi}{dt} &= \frac{\mu_0 I}{2\pi} \ln \Biggr( \frac{d-r}{r} \Biggr) \frac{dx}{dt}\\
&= \frac{\mu_0 I}{2\pi} \ln \Biggr( \frac{d-r}{r} \Biggr) v
\end{align*}
[/itex]

So does the above look correct?

The part I am unsure of is how to express that we can ignore the flux from the sliding bar. Especially considering no specific relation is given between x and d.

Thanks in advance.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,622
3,016
Hello, DiogenesTorch. Welcome to PF!

The magnitude of the magnetic field of half infinite wire is [itex] |\vec{B}|= \frac{\mu_0 I}{4\pi s} [/itex] where [itex]s[/itex] is the perpendicular distance from the wire.
This is not correct, except at points directly above or below the finite end of the wire. At other points near the finite end, the field produced by the wire would be more complicated. However, for points way down the wire (in the x direction) so that the wire extends far in both directions from the point, the half-wire would produce a field approximately the same as a wire that extends infinitely far in both directions.

The part I am unsure of is how to express that we can ignore the flux from the sliding bar. Especially considering no specific relation is given between x and d.
You are only concerned with the change in flux during a time dt. So, you should think about the change in flux as the rod moves a distance dx during the time dt and the rod is very far from the finite ends of the wires. Try to see why the current in the rod doesn't need to be taken into account when considering the change in flux through the circuit.
 
Last edited:

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