Two long, parallel wires carry currents of I 1 = 2.96 A and I 2 = 5.05 A in the

Question

Two long, parallel wires carry currents of I1 = 2.96 A and I2 = 5.05 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).

magnitude ________?T

direction________ ° from the positive x-axis

(b) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 5.05-A current.

magnitude ________?T

direction________ ° to the left of the vertical

Explanation / Answer

Magnetic field due to a wire is given by

B=uoI/2pi*r

a)

At the midpoint ,the Magnetic field due to a each wire is parallel to the y-axis ,so net field is given by

Bnet=B1y-B2y= uo(I1-I2)/2pir

Bnet=(4pi*10-7)(2.96-5.05)/2pi*0.1

Bnet=-4.18 uT

So magnitude

|Bnet|=4.18 uT

Direction is towards the bottom of the page i.e 270o  from the positive x-axis

b)

Wire 1

B1=uoI1/2pi*r1=(4pi*10-7)*2.96/2pi*sqrt[0.22+0.22]

B1=2.09 uT

Wire 2

B2=uoI2/2pi*r2=(4pi*10-7)*5.05/2pi*0.2

B1=5.05 uT

X-Component of Magnetic field

Bx=B1Cos135 + B2Cos180=2.09Cos135+5.05Cos180

Bx=-6.54 uT

Y-Component of Magnetic field

By=B1Sin135 + B2SIn180=2.09Sin135+5.05Sin180

By=1.48 uT

Magnitude

B=sqrt[Bx2+By2]=6.7 uT

Direction

o=tan-1|-6.54/1.48|=77.25o to the left of the vertical

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