Two long, parallel wires carry currents of I 1 = 2.72 A and I 2 = 5.35 A in the
ID: 1731393 • Letter: T
Question
Two long, parallel wires carry currents of I1 = 2.72 A and I2 = 5.35 A in the direction indicated in the figure below. (Choose the line running from wire 1 to wire 2 as the positive x-axis and the line running upward from wire 1 as the positive y-axis.)
(a) Find the magnitude and direction of the magnetic field at a point midway between the wires (d = 20.0 cm).
(b) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 5.35-A current.
Explanation / Answer
A).
Net magnetic field can be determined by the vector sum of magnitudes of magnetic fields due to individual wires
B= (uo/4*pi)* 2/r* (i2- i1)
= 10^-7* 2* (5.35-2.72)/0.1
= 5.26 uT
Directed 270° from +x axis
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Angle, x= arctan(d/d)= 45
Magnetic field at P due to i1
B1= 10^-7*2*2.72/(0.2* sqrt(2))* ( - cos(90 -x)i + sin (90-x)j)
= ( - 1.36 i + 1.36 j)*10^-6
Magnetic field due to P due to i2
B2= - 10^-7*2*5.35/0.2 i = - 5.35*10^-6 i
Net magnetic field
B= [- (5.35+1.36)i +1.36 j]*10^-6
= [- 6.71i + 1.36 j] uT
Magnitude, B= 6.846 uT
Direction, theta= 180 - arctan(1.36/6.71)
Theta= 168.542
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Comment in case any doubt.. good luck
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