As shown in the figure below, two long parallel wires (1 and 2) carry currents o
ID: 1262370 • Letter: A
Question
As shown in the figure below, two long parallel wires (1 and 2) carry currents of I1 = 2.74 A and I2 = 4.50 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.
_____ ?T
magnitude_____ ?T
direction ______ ? counterclockwise from the +x-axis As shown in the figure below, two long parallel wires (1 and 2) carry currents of I1 = 2.74 A and I2 = 4.50 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 _____ µT direction ______ ° counterclockwise from the +x-axisExplanation / Answer
magnetic field due to current carrying wire
B= (?0I)/2?d
here ?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)(2.74A) / 2?(0.10 m)
= 5.48 *10^-6 T
If you use the right hand rule I explained at point p its direction will be directed along the -x axis (180 degrees)
From Pythagoreans theorem, the distance second wire and point p is
sqrt(10^2 + 10^2) = 14.421 cm = 0.1441m
So B = (4?*10^-7)(4.5 A) / 2?(0.144 m) = 6.2*10^-6 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
5.48*10^-6 Tcos180 = -5.48*10^-6 T
5.48*10^-6 Tsin180 = 0
6.2*10^-6 Tcos45 = 4.38*10^-6 T
6.2*10^-5 Tsin45 = 4.38*10^-6
--------------------------------------...
sum of x components = -1.09*10^-6
sum of y components = 4.38*10^-6
Magnitude of magnetic field is
B = sqrt ( ( -1.09*10^-6)^2 + ( 4.38*10^-6 )^2)
= 4.515 micro T
directon = - 76 degrees = 280 degrees from positive x-axis.
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