# Induced Current

This problem is difficult to describe, so I'll post a picture.
http://img71.imageshack.us/my.php?image=pic1ik.gif

The figure above shows a rod of length L caused to move at a constant speed v along horizontal conducting rails. The magnetic field B (the magnitude and direction of which are qualitatively shown by the figure) is not constant, but is supplied by a long wire parallel to the conducting rails. This wire is a distance a from the rail and has a current i.

L=3.13 cm, v=3.11 m/s, a=15.6 mm, and i=11 A.

What is the induced emf (e) in the rod?

---
B = (u_0 I)/(2pi y), and it is not uniform, so I integrated over y=a...L
I got (u_0 I)/(2pi)*ln(L/a).

emf = vBL = v * (u_0 I)/(2pi)*ln(L/a) * L = 3.11 * 1.544597242E-6 * 0.0313 = 1.503557293E-7 V

That is not the right answer, however.
I double-checked that my calculations are correct. So I'm guessing that my steps are incorrect. Can anyone point me to where I'm going wrong?

Thanks.

Related Introductory Physics Homework Help News on Phys.org
Hootenanny
Staff Emeritus
Gold Member
I would try this;

$$B = \frac{\mu_{0} I}{2\pi y}$$

a is a constant; $y = a + L$

$$\frac{dB}{dL} = \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL$$

$$B = \int^{0.0313}_{0} \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL$$

See if that works.

~H

Last edited:
Päällikkö
Homework Helper
Shouldn't the integration be done over y=a..a+L?

EDIT: I'm always too slow .

B was approximately 3.326E-4, and emf was 3.23762818E-5 V, which was still incorrect.
Am I using the right L for emf = vBL? I'm using 3.13 cm (0.0313 m).

Hootenanny
Staff Emeritus
Gold Member
I think you may have integrated in correctly you should obtain;

$$B = \int^{0.0313}_{0} \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL$$

$$B = \left[ \frac{1}{2}\mu_{0}I \log (a + L) \right]^{0.0313}_{0}$$

Also ensure that you are converting correctly, note that a is given in mm. You are using the correct equation here;

nahya said:
Am I using the right L for emf = vBL? I'm using 3.13 cm (0.0313 m)
~H

Last edited:
Thanks, that worked.
Weird, though, because I had used my calculator to graph the equation and integrate graphically.

Hootenanny
Staff Emeritus
No problem. I prefer pen and paper 