Biot-Savart Law, right angle

[ParoXsitiC]

Homework Statement

http://i.minus.com/1333003834/Q661ScjxBUxkrL2FfmVFPQ/iTEfM3UTAVtTa.png

Homework Equations

B = ∫ ([μ0 / 4pi] * I * ds-vector x r-hat) / r^2

The Attempt at a Solution

I know the horizontal line will not add anything to the magnetic field (B), so focusing on the vertical line.

I take a little bit of length (ds) which I will call dy.

dy x r-hat = dy sin θ
r = sqrt(x^2+y^2)
sin θ = x / r

do all your substitutions and get:

B = ([μ0 / 4pi] * I * x ) ∫ dy / (x^2+y^2)^(3/2)

At this point I am confused on my limits of integration, I know for an infinite long straight wire I use -∞ to ∞.

In my notes I have an example where it goes from -y1 to y2 and comes out with

B = ([μ0 / 4pi] * I ) / x * (cos θ1 - cos θ2)

where θ1 is the angle between -y1 and the point i am finding, and θ2 is 180 - θ1.
This whole θ thing is tripping me up, how did it get there ( I am assuming trig subsitutation). Further more how can I get a grasp on what θ1 would be?

I guess -y1 in my situation is just y, and y2 is 0.

so θ1 = inverse-tan (x/y) and thus θ2 = 180 - inverse-tan (x/y)?

 

[tiny-tim]

Hi ParoXsitiC!

(try using the X² and X₂ buttons just above the Reply box )

(and write arctan or tan^-1, not inverse-tan)

(You're rambling a bit , so I won't answer point-by-point)

θ isn't a substitutuion, it's the actual angle, between the current and the line from the point to P

You can either do ∫ dy, in which case your limits are the endvalues of y, in this case 0 and ∞

or you can do ∫ dθ, in which case your limits are the endvalues of θ, in this case 0 and π/2.

(btw, can't you just say it's half the value for a whole line?)

 

[ParoXsitiC]

That makes sense but how did they get to the two cosines mathematically?

I guess my understand is that the formula is a simplified equation that does not include an integral, but I don't understand how it was derived, nor do I understand which angle is taken for θ₂

 

[tiny-tim]

That makes sense but how did they get to the two cosines mathematically?

oh, cos = adj/hyp (adjacent/hypotenuse)

= y/√(x² + y²) ​

(and θ₂ is the angle PYO, where Y = (0,y), as y -> ∞)

 

[asteriatic]

I would like to clarify that the Biot-Savart Law is a fundamental law in electromagnetism that describes the magnetic field created by a steady current in a wire. It states that the magnetic field at a point is directly proportional to the current and the vector product of the current element and the displacement vector from the current element to the point, and inversely proportional to the square of the distance between them.

In this specific problem, we are looking at a right angle configuration where one side of the wire is horizontal and the other is vertical. As correctly noted, the horizontal line will not contribute to the magnetic field since the cross product of the current element and the displacement vector will be zero. Therefore, we only need to focus on the vertical line.

To find the magnetic field at a point due to this vertical line, we can use the Biot-Savart Law and integrate over the length of the wire. However, as pointed out, there is some confusion about the limits of integration and the use of trigonometric functions. In this case, it would be helpful to draw a diagram to visualize the situation.

From the diagram, we can see that the displacement vector from the current element (dy) to the point (P) can be represented by r, which is the hypotenuse of a right triangle. Using trigonometry, we can express the angle θ between the displacement vector and the horizontal line as θ = arctan(x/y).

Now, to find the magnetic field at point P, we need to integrate over the length of the wire (from y=0 to y=h) using the Biot-Savart Law. This will give us the following expression:

B = [μ0 / 4π] * I * x * ∫ dy / (x^2 + y^2)^(3/2)

Integrating this expression will give us the desired result:

B = [μ0 / 4π] * I / x * (cos θ1 - cos θ2)

where θ1 is the angle between the displacement vector and the vertical line at y=0, and θ2 is the angle between the displacement vector and the vertical line at y=h. Using trigonometric identities, we can express these angles as:

θ1 = arctan(x/h) and θ2 = arctan(x/0) = π/2