Creating a Zero-Magnetic Field w/ Atom Chip: Current, Resistance & Thickness

[tim_3491]

Show that if a straight wire carrying current I is oriented at right angles to a uniform external magnetic field B, there will be a region where the magnetic field is zero. If the wire is actually ‘printed’ onto the surface of an ‘atom chip’ and the atoms are to be trapped 500 micrometre from the surface using an external field of 25 x 10^-4 T, what current will the wire need to carry? If the resistance of the wire, which is 6mm long and 150µm wide is not to exceed 0.1
ohm, how thick must it be if made of copper?
(See http://www.physics.uq.edu.au/atomoptics/atomchip.html for the actual device we made. The potential energy of atoms in one spin state is zero where the field is zero, so they can be trapped there.

Anyone have any ideas, we have recently been doing emf and electric fields

Not to sure even on how to start it

 

[jbsteele]

. Answer: In order for a region of zero magnetic field to exist, the current I must be oriented so that it creates a magnetic field in the opposite direction to the external field B. This is known as the Lorentz force law. For the atoms to be trapped 500 micrometres from the surface, the current I must be strong enough to counteract the external field of 25 x 10^-4 T. This can be calculated using the formula I=B/μ0, where μ0 is the permeability of free space (4π*10^-7 Tm/A). Substituting the given values gives I = 6.28*10^-2 A. The resistance of the wire can be calculated using the equation R = ρl/A, where ρ is the resistivity of the material and l and A are the length and cross-sectional area of the wire. For copper, the resistivity is 1.68*10^-8 Ωm. Substituting the given values gives R = 0.11 Ω. This is slightly above the desired resistance of 0.1 Ω, so the thickness of the wire must be increased to reduce the resistance. Using the equation A = lt, where l and t are the length and thickness of the wire respectively, the thickness of the wire can be determined. Substituting the given values gives t = 8.8*10^-5 m (88 µm). Therefore, the wire must be 88 µm thick in order to have a resistance of 0.1 Ω.

 

[Moetasim]

Sure, I can help you with this. Let's break down the problem step by step.

First, let's consider the concept of a zero-magnetic field. This means that there is no magnetic field present in a certain region. In order to achieve this, we need to have the external magnetic field B and the magnetic field produced by the wire, which we will call B_wire, cancel each other out. This can be achieved when B and B_wire are equal in magnitude but opposite in direction.

Next, let's think about the orientation of the wire. We are told that the wire is oriented at right angles (perpendicular) to the external magnetic field B. This means that the direction of B_wire will be parallel to the direction of the current I flowing through the wire. This is known as the right-hand rule for magnetic fields.

Now, let's consider the equation for the magnetic field produced by a straight wire:

B_wire = μ0I/2πr

Where μ0 is the permeability of free space, I is the current flowing through the wire, and r is the distance from the wire. We can see that the magnetic field is inversely proportional to the distance from the wire, so as we move away from the wire, the magnetic field decreases.

So, in order to have a region of zero-magnetic field, we need to have the magnetic field produced by the wire, B_wire, equal to the external magnetic field B at a certain distance from the wire. This means that we can set the two equations equal to each other:

B_wire = μ0I/2πr = B

Solving for r, we get:

r = μ0I/2πB

This gives us the distance from the wire at which the magnetic field will be zero.

Now, let's move on to the second part of the problem. We are told that the atoms need to be trapped 500 micrometers from the surface using an external field of 25 x 10^-4 T. We can use the equation we just found to calculate the current needed in the wire:

I = 2πBr/μ0 = 2π(25 x 10^-4 T)(500 x 10^-6 m)/μ0 = 0.0395 A

So, the wire will need to carry a current of 0.0395 A to create a zero-magnetic field at a distance of 500