Recent content by scoldham

Taking into account that my value for r is incorrect, is this the correct approach? If I plugged in the correct value for r, this would work? Thanks for all the assistance so far!

So, generically speaking, r = \sqrt{s^2 + <x,y>^2} where <x,y> will change depending on which segment I'm considering. I could integrate, B(<x,y>) = \frac{\mu_0 I}{4 \pi} \int^{a}_{0} \frac{ds}{|r|^2} using the generic value for r which will produce an equation that gives B as a function of...

Does r = \sqrt{s^2 + \frac{a^2}{16}}? --edit-- Correction: r = \sqrt{s^2 + \frac{x^2}{16}}

If I'm understanding correctly, B(\hat r) = \int \frac{\mu_0 I}{4 \pi} \frac{dl \times \hat r}{|r|^2} dl = ds \hat r = <x,y> Thus, B(<x,y>) = \frac{\mu_0 I}{4 \pi} \int \frac{ds \times \hat r}{|<x,y>|^2} ds \times \hat r = ds Integrate from 0 to a, B(x,y) = \frac{\mu_0 I}{4 \pi}...

I not sure what value to use for r in Biot-savart. Could I use a double integral and integrate over r as well considering that it is a changing value?

Homework Statement What is the magnitude and direction of magnetic field at point P shown in the figure attached if i = 10 A and a = 8.0 cm? Express the answer in vector notation. Homework Equations B = \frac{\mu_0I}{2\pi d} The Attempt at a Solution I'm really not sure how to...

I'm not sure I understand, Are you saying that q_1 E_{net} = v_1 and that the same is true for q2 and v2... so, v_1+v_2 =\Delta V Therefore w = Q \Delta V

Homework Statement Referring to the figure attached, how much work must be done to bring a particle, of charge Q = +16e and initially at rest, along the dashed line from infinity to the indicated point near two fixed particles of charges q1 = +4e and q2 = -2e? Distance d = 1.40 cm, theta...

That was the problem.. I was using the wrong values for the constants... I'll work it out and let you know if I have any more problems. Thanks for the help!

I got it at this point. Thanks for all your help. And just for the record.. you were picturing it right.

z_2 = 3.00 meters

http://www.wolframalpha.com/input/?i=int%28int%28-3b+dy%2C+1%2C+0%29dx%2C+1%2C+4+%29 http://www.wolframalpha.com/input/?i=int%28int%283b+dy%2C+1%2C+0%29dx%2C+1%2C+4+%29

Something isn't working out. Each side cancels the other side... including the sides which involve b. Looking at the left face, d \vec A = -i dy dz and \vec E \cdot d \vec A = -(10.00 + 2.00x) dy dz Integrating the left and right face each over z_1 to z_2 and 0 to y_2 gives 48 and...

So... I need to determine d \vec A and dot it with \vec E for each face of the shape. Integrate to remove the differentials, then sum the result of each integration to solve for b? Am I finally on the right track? Thank you so much for your help so far.

So... square the i and j components, sum them, and take the square root?