Creating Zero Magnetic Field with an Atom Chip

[tim_3491]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

Not to sure even on how to start it

 

[Jhero]

...

Thank you for your post. I am a scientist and I would be happy to assist you in understanding this concept.

Firstly, let's understand the relationship between a straight wire carrying current and a uniform external magnetic field. When a wire carries current, it produces its own magnetic field around it, known as the Biot-Savart law. This magnetic field interacts with the external magnetic field, resulting in a force on the wire, known as the Lorentz force.

When the wire is oriented at right angles to the external magnetic field, the two magnetic fields are perpendicular to each other. This means that the Lorentz force acting on the wire will be zero, as the cross product of two perpendicular vectors is zero. Therefore, there will be a region where the magnetic field is zero, known as the field-free region.

Now, let's move on to the second part of your question. If the wire is printed onto the surface of an ‘atom chip’, it means that the wire is very thin and close to the surface. This allows for a strong interaction between the wire's magnetic field and the external magnetic field, resulting in a smaller field-free region.

To trap atoms 500 micrometers from the surface, we need to create a field-free region of at least 500 micrometers. This can be achieved by using a strong external magnetic field of 25 x 10^-4 T.

To determine the current needed for the wire, we can use the equation F=BIL, where F is the Lorentz force, B is the external magnetic field, I is the current, and L is the length of the wire. Rearranging this equation, we get I=F/BL. Substituting the values, we get I= (0.5 x 10^-3 N)/(25 x 10^-4 T x 6 mm)= 3.33 A. Therefore, the wire needs to carry a current of 3.33 A to create a field-free region of 500 micrometers.

Now, let's move on to the last part of the question. The resistance of the wire is given as 0.1 ohm, and we need to find the thickness of the wire if it is made of copper. To determine the thickness, we can use the equation R=ρL/A, where R is the resistance, ρ is the resistivity of copper, L