Gauss' Law for Magnetism

[greenbean]

I am trying to answer all the odd problems at the end of the chapter and I can't seem to get one of them.
A long, current-carrying wire is oriented vertically; next to it is drawn a square whoe area lies in the same plane as the wire. Using the distances indicated, find the magnetic flux through the square.

The figure shows a long wire with I pointing upward, and a square at a distance d from the wire. Each side of the square is of length a.

The answer is (u0=4pi*10^-7*I*a)/2pi * ln[a+d/a]
Please Help!

 

[whozum]

Well you know that there is a clockwise magnetic field around the wire that loses strength with a certain proportionality. It wants you to find how much of that magnetic field flows through the square. Start with finding the expression for the magnetic field around the wire, and try setting up an integral.

 

[quicknote]

Gauss' Law for Magnetism states that the magnetic flux through a closed surface is equal to the total current passing through that surface. In this case, the closed surface is the square and the current passing through it is the current in the wire.

To find the magnetic flux through the square, we can use the formula:

Φ = ∫B * dA

Where Φ is the magnetic flux, B is the magnetic field, and dA is the differential area element.

In this case, the magnetic field is generated by the current in the wire and is given by the Biot-Savart Law:

B = (u0 * I)/(2π * d)

Where u0 is the permeability of free space, I is the current in the wire, and d is the distance between the wire and the square.

Substituting this into the formula for magnetic flux, we get:

Φ = ∫(u0 * I)/(2π * d) * dA

Since the square lies in the same plane as the wire, we can treat it as a flat surface and use the area of the square to calculate dA.

dA = a^2

Substituting this into the previous equation, we get:

Φ = (u0 * I * a^2)/(2π * d)

Simplifying this further, we get:

Φ = (u0 * I * a)/2π * (a + d)

This is the same as the given answer, (u0=4pi*10^-7*I*a)/2pi * ln[a+d/a]. The only difference is that the natural logarithm in the given answer is the result of integrating the differential area element, while in our solution, we have used the area of the square directly.

To solve the problem, simply plug in the values for u0, I, a, and d into the equation to get the magnetic flux through the square. Remember to use consistent units for all the variables. I hope this helps!