Electricity and Magnetism Question. Current related.

[Partap03]

Okay, I will be honest, this is a homework assignment. For that matter this is the last homework assignment which is very crucial to preparing for the final. But the problem is I have 2 more finals for which I need to study for. So if you can please help me out and give a detailed solution I will appreciate it. Thanks in advance.


A rectangular loop of wire (sides a and b) lies flat in the x-y plane centered at the origin and carries current I in the clockwise direction (viewed from above). Find the magnetic field (direction and magnitude) at the origin.

 

[jbsteele]

The magnetic field at the origin due to the rectangular loop of wire is given by the equationB = μ_0I/(2πab) * [cos^-1(b/a)+sin^-1(b/a)].Where μ_0 is the permeability of free space, I is the current, a and b are the sides of the rectangle, and [cos^-1(b/a)+sin^-1(b/a)] is the angle between the two sides. The magnitude of the magnetic field is therefore B = μ_0I/(2πab) * [cos^-1(b/a)+sin^-1(b/a)] and the direction is out of the plane of the rectangle.

 

[nlightner]

Sure, I would be happy to help you with this problem. The first step in finding the magnetic field at the origin is to use the right-hand rule to determine the direction of the magnetic field around a current-carrying wire. If you point your thumb in the direction of the current, your fingers will curl in the direction of the magnetic field.

In this case, since the current is clockwise, the magnetic field will be counterclockwise around the wire. Now, let's consider the magnetic field at the origin due to each side of the rectangular loop separately.

For the top and bottom sides of the loop, the magnetic field will be perpendicular to the wire and will point in the same direction for both sides (out of the page). This is because the current is flowing in the same direction on both sides.

For the left and right sides of the loop, the magnetic field will be parallel to the wire and will point in opposite directions for each side. This is because the current is flowing in opposite directions on these sides.

Now, to find the total magnetic field at the origin, we need to add the contributions from each side of the loop. Since the top and bottom sides have the same magnitude and direction, their contributions will cancel out. However, the left and right sides have the same magnitude but opposite directions, so their contributions will add up.

Using the Biot-Savart law, we can calculate the magnitude of the magnetic field at the origin due to one side of the loop:

B = μ0I/2πr

Where μ0 is the permeability of free space, I is the current, and r is the distance from the wire to the point of interest (in this case, the origin). Since the distance from the wire to the origin is a/2, the magnitude of the magnetic field at the origin due to one side of the loop is:

B = μ0I/4πa

Since there are two sides contributing to the magnetic field at the origin, the total magnitude of the magnetic field at the origin is:

Btotal = 2(μ0I/4πa) = μ0I/2πa

Finally, we know that the direction of the magnetic field at the origin will be perpendicular to the plane of the loop, so it will be in the z-direction. Therefore, the complete solution for the magnetic field at the origin is:

B = μ0I/2πa in the