As shown in the figure above, two long parallel wires (1 and 2) carry currents o
ID: 1262501 • Letter: A
Question
As shown in the figure above, two long parallel wires (1 and 2) carry currents of I1 = 2.76 A and I2 = 5.00 A in the direction indicated.Determine the magnitude and direction of the magnetic field at point P, located d = 10.0 cm above wire 1.
magnitude
direction in degrees counterclockwise from the +x-axis
As shown in the figure above, two long parallel wires (1 and 2) carry currents of I1 = 2.76 A and I2 = 5.00 A in the direction indicated.Determine the magnitude and direction of the magnetic field at point P, located d = 10.0 cm above wire 1. magnitude direction in degrees counterclockwise from the +x-axisExplanation / Answer
For a current carrying wire your thumb is the direction of the current and your fingers curled is the magnetic field direction. To determine the actual direction at a certain point in time you go to the desired point and go off on a tangent (which is just like going off at a right angle).
To actually solve for the magnetic field from one wire you use:
B= (?0I)/2?d and ?0 is a constant which equals 4?*10^-7 N/A^2
So from the wire on the left coming out of the page B = (4?*10^-7)(2.76A) / 2?(0.1m) = .552*10^-5 T
(I learned the units of a magnetic field as Tesla I'm not quite sure what B is sorry :( )
If you use the right hand rule I explained at point p its direction will be directed along the -x axis (180 degrees)
We don't know d for the second wire but we can use Pythagoreans theorem to solve for it.
sqrt(10^2 + 10^2) = 14.14cm = 0.1414m
So B = (4?*10^-7)(5A) / 2?(0.1414m) = 0.7072*10^-5 T
Since this is a nice triangle we know the angles in the triangle are 45-45-90 so using the right hand rule the direction is 45 degrees from the +x axis.
Now we can set up a vector table to solve for the net magnetic field
-.552*10^-5 Tcos180 = -552*10^-5 T 552*10^-5 Tsin180 = 0
0.7072*10^-5 Tcos45 = 5*10^-6 T0.7072*10^-5 Tsin45 = 5*10^-6
--------------------------------------...
sum of x components = -5.52*10^-6
sum of y components = 10*10^-6
Answer:
net B = sqrt( (-10*10^-6)^2 + (5.52*10^-6)^2 ) = 11.423*10^-6 T = 11.423
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