Net magnetic field caused by current from two wires

[mm2424]

Homework Statement

The attached figure shows, in cross section, two long parallel wires that are separated by distance d = 18.6 cm. Each carries 4.34 A, out of the page in wire 1 and into the page in wire 2. In unit-vector notation, what is the net magnetic field at point P at distance R = 34.2 cm, due to the two currents?

Homework Equations

B = iμ_o/2∏R

The Attempt at a Solution

(1) The magnetic field lines around wire have a counterclockwise direction. The magnetic field lines around wire 2 have a clockwise direction. When the field lines hit point P, those from wire 1 point to the top right while those from wire 2 point to the bottom right.

(2) Angle a in the diagram is equal to the tan^-1 of 9.3/34.2. Therefore, it = 15.21. This is also the measure of the angle (measured from the x axis) of the tangent of the magnetic field lines at P from both wire 1 and wire 2.

(3) Bnet = 2[iμ_o/2∏R] = 4.79 x 10^-6 T.

(4) However, only the x components of the magnetic fields survive and add. The y components cancel. Therefore, we must multiply the Bnet by cos(15.21), which yields 4.62 x 10^-6 T.

(5) Because the total net field points to the right, the answer is (4.62 x 10^-6 T)(i hat)

Is this right? I can't find the answer to this question anywhere. Thanks!

 

[Redbelly98]

You mostly have the right idea, but I found two problems.

(2) Angle a in the diagram is equal to the tan^-1 of 9.3/34.2. Therefore, it = 15.21. This is also the measure of the angle (measured from the x axis) of the tangent of the magnetic field lines at P from both wire 1 and wire 2.

You're correct about angle a, but think more carefully about what angle B makes. It might help to think about a point P₂ that is much farther away, where the angle a₂ would be very small.

(3) Bnet = 2[iμ_o/2∏R] = 4.79 x 10^-6 T.

I get something different, but not far off from what you get for Bnet. What are you using for R? You should include a calculation of R in your work.

Also it's really not proper to call this Bnet, which would include the cos(15.21) factor you do later on. I recommend not including the factor of 2, and call this a calculation of B due to one wire.

(4) However, only the x components of the magnetic fields survive and add. The y components cancel. Therefore, we must multiply the Bnet by cos(15.21), which yields 4.62 x 10^-6 T.

(5) Because the total net field points to the right, the answer is (4.62 x 10^-6 T)(i hat)

Is this right? I can't find the answer to this question anywhere. Thanks!

Mostly right, just rethink about the angle for B, and show your calculation of R.

 

[mm2424]

Thanks! Is there a method for figuring out the angle B makes? I was fine when it came to finding the correct angles in kinematics, but I consistently choose the wrong angle in magnetism questions. There might be some geometry rule I've forgotten or something.

 

[Redbelly98]

The method at play is the basic geometry rules concerning angle measures. Also, realizing that a magnetic field line forms a circle with the wire at the center. That means B (due to wire 1) at point P is at a right angle to the red line in your figure.

 

[asteriatic]

Your solution looks correct. The net magnetic field at point P due to the two currents is indeed (4.62 x 10^-6 T)(i hat). Good job using the right-hand rule to determine the direction of the magnetic field lines and considering the x and y components separately. Keep up the good work!