Calculating Magnetic Field at a Point Using Biot-Savart Law

AI Thread Summary
To calculate the magnetic field at point O due to a long wire and a quarter circular arc using the Biot-Savart Law, the equation B = μ0 * I / 2a is referenced, but it may not yield the correct result if only the arc is considered. The integral form of the Biot-Savart Law should be applied, focusing on the arc's contribution while recognizing that the current and distance are constant. The relationship between the arc and the point O must be examined, particularly how the chord connecting O to the arc is perpendicular to the arc's infinitesimal length. An additional factor may be necessary in the calculations. It's also important to verify unit consistency throughout the solution process.
Eva01
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Homework Statement



A long (effectively infinite) wire segment
is connected to a quarter of a circular arc with
radius a. The other end of the arc is connected
to another long horizontal wire segment. The
current is flowing from the top coming down
vertically and flows to the right along the pos-
itive x-axis. I have included the image below.
I = 7.5A, a = 0.69 m. Find the magnetic field at O?

Homework Equations



B = μ0 * I / 2a

The Attempt at a Solution



As far as I understand since we only consider the arc part then I use the formula from above but I am not getting the right answer.

B = 4∏ * 10^-7 * 7.5 / 2*0.69 = 6.8295 N/m
 

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Hello Eva01,

Firstly you need to consider the origin of your equation :
B = μ0 * I / 2a

which is the Biot Savart law.

B = \frac{\mu_{0}}{4 \pi}\int \frac{(I dl \times \hat{r})}{ |\vec{r}|^{2}}

For your system you do only need to use the arc system but you should find out why..

For the arc itself, consider the law above. Clearly the I and the denominator are constant. So we need to solve

\int (dl \times \hat{r})

Remembering that a chord (r) connecting the point O to the arc will always be perpendicular to an infinitesimal length of the arc. This should give you a similar equation to what you have been using but it requires an additional factor somewhere.


I would also advise that you check your units at some point.

Hope this helps
 
Thank you for your help!
 
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