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Two long, straight wires cross each other at right angles, as shown in the figur

ID: 1423982 • Letter: T

Question

Two long, straight wires cross each other at right angles, as shown in the figure below. (Assume I1 = 5.05 A and I2 = 3.16 A.)

(a) Find the direction and magnitude of the magnetic field at point P, which is in the same plane as the two wires.

magnitude = ______ T

direction = out of the page

(b) Find the magnetic field at a point 30.0 cm above the point of intersection (30.0 cm out of the page, toward you).

magnitude = _______ T direction = _______    ° counterclockwise from the +x-axis, parallel to the xy-plane

Explanation / Answer

  I'm going to assume that the currents are flowing in the positive x and positive y directions...
a) Use the right hand rule. The x wire's field at point P goes out of the page, and the y-wire field goes into the page.
Field due to x wire:
B = ui/2r = 2E-7i/r = 2E-7*5.05/0.4 = 2.53E-6
Field due to y wire:
B = 2E-7*3.16/0.3 = 2.1E-6
x wins. Net magnitude: 2.53E-6 - 2.1E-6 = 4.3 E-7 T
Since x wins, the field goes out of the page.

b) I'm going to call that point Z
Distance from x wire to point Z: (0.30²+0.4²) = 0.5 m
Plug into formula
B = 2E-7*5.05/0.5 = 2.02E-6 T
The direction is a little bit tricky. It will be pointing out of the page and towards you. I'm going to call the positive x axis east, positive y axis north, and out of the page up. So the angle would be

arctan(0.4/0.30) = 53.13 [S 53.13 UP]

Do the same thing with y.
(0.30²+0.3²) = 0.424 m
B = 2E-7*3.16/0.424 = 1.49E-6 T
angle: arctan(0.3/0.3) = 45. [E 45 DOWN]
So now combine 2.02E-6 [S53.13 U] and 1.49E-6 [E45D] using components.
In the vertical direction: 5.3E-7 up

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