Total Josephson current through junction with magnetic field

In summary, the integral is over the area where ##x## and ##z## range from ##0## to ##d_x## and ##y## ranges from ##-d_y/2## to ##d_y/2##.
  • #1
Mr_Allod
42
16
Homework Statement
Show that the total Josephson current though a Josephson Junction is:
$$d_xd_yj_0 \frac {\phi_0}{2\pi \phi} \left[ \cos(\psi_0) - \cos(\psi_0 + \frac{2\pi \phi}{\phi_0}) \right]$$
Relevant Equations
Total flux through the junction: ##\phi = Bd_xd_y##
Current density at position x: ##J(x) = j_0\sin(\psi_0 + \frac{2\pi x}{\phi_0}Bd_y)##
Hello there,

I am given a diagram of a Josephson Junction like so:

JJ.PNG

With a magnetic field ##B = \mu_oH## in the z-direction. I'm reasonably sure ##d_x,d_y,d_z## are normal lengths, not infinitesimal lengths although that is up for debate. Using the above equations I rearrange the expression for J(x) to be in terms of ##\phi## and compute the following integral:

$$\int_{\frac {-d_x}{2}}^{\frac {d_x}{2}}\int_{\frac {-d_y}{2}}^{\frac {d_y}{2}} j_0\sin(\psi_0 + \frac{2\pi \phi}{\phi_0d_x}x)d_xd_y$$

This yielded:

$$d_xd_yj_0 \frac {\phi_0}{2\pi \phi} \left[ \cos(\psi_0 - \frac{\pi \phi}{\phi_0}) - \cos(\psi_0 + \frac{\pi \phi}{\phi_0}) \right]$$

Where ##\psi_0## is the phase difference at ##x=0##. This is not the same as the expression we were given in the problem but it is quite similar which makes me think I am on the right track. Is there something else I need to do with the expression we are given before integrating to find the total current?
 
Physics news on Phys.org
  • #2
Mr_Allod said:
$$\int_{\frac {-d_x}{2}}^{\frac {d_x}{2}}\int_{\frac {-d_y}{2}}^{\frac {d_y}{2}} j_0\sin(\psi_0 + \frac{2\pi \phi}{\phi_0d_x}x)d_xd_y$$

Shouldn't the integration be over ##x## and ##z## rather than ##x## and ##y##?

Maybe the origin of the coordinate system is such that ##x## ranges from ##0## to ##d_x## rather than from ##-d_x/2## to ##d_x/2## and ##z## ranges from ##0## to ##d_z## rather than from ##-d_z/2## to ##d_z/2##.
 
  • #3
TSny said:
Shouldn't the integration be over ##x## and ##z## rather than ##x## and ##y##?

Maybe the origin of the coordinate system is such that ##x## ranges from ##0## to ##d_x## rather than from ##-d_x/2## to ##d_x/2## and ##z## ranges from ##0## to ##d_z## rather than from ##-d_z/2## to ##d_z/2##.
I think you're right on both counts. I suppose if I had paid more attention to the figure I would have realized I'm using the wrong origin. Integrating from ##0## to ##d_{x,z}## gives the right answer. With regards to the area of integration, I think I was just forcing the solution to be in line with the expression we're given and it's entirely possible there was a typo in it. It wouldn't be the first time.
 
  • Like
Likes TSny

Related to Total Josephson current through junction with magnetic field

1. What is the Josephson effect?

The Josephson effect is a phenomenon in which a supercurrent (a current of paired electrons) can flow through a weak link between two superconductors. It was first discovered in 1962 by Brian Josephson and has since been studied extensively in the field of superconductivity.

2. How does a magnetic field affect the Josephson current?

A magnetic field can cause a phase difference between the superconducting wavefunctions in the two superconductors, which in turn affects the Josephson current. The total Josephson current through a junction with a magnetic field depends on the strength and orientation of the magnetic field.

3. What is the critical current in a Josephson junction?

The critical current is the maximum current that can flow through a Josephson junction without causing a voltage drop. It is directly related to the superconducting gap energy and the strength of the coupling between the two superconductors in the junction.

4. How can the Josephson current be manipulated?

The Josephson current can be manipulated by changing the temperature, magnetic field, or voltage applied to the junction. It can also be controlled by varying the thickness or material of the weak link between the two superconductors.

5. What applications does the Josephson effect have?

The Josephson effect has many practical applications, including superconducting quantum interference devices (SQUIDs) used in sensitive magnetic field measurements, Josephson voltage standards for precise voltage measurements, and superconducting quantum computing. It also has potential uses in high-speed electronics and energy-efficient power transmission.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
1
Views
854
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
391
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
2
Replies
46
Views
5K
Back
Top