Two long, parallel wires carry currents of I1 = 3.10 A and I2 = 5.35 A in the di

Question

Two long, parallel wires carry currents of I1 = 3.10 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).

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 5.35-A current.

magnitude ?T

direction ° to the left of the vertical

Explanation / Answer

a) current is coming out of the page

at mid point between them magnetic field will be subtractive in nature

Bnet = (Uo*i1/2*pi*0.8d) - (Uo*i2/2*pi*0.5d) = Uo*(i1 - i2)/(2*pi*0.5d) = 4*pi*10^-7*(3.10 -5.35)/(2*pi*0.5*0.20)

=>magnitude of Bnet = 4.5*10^-6T

direction is -90 degree from x-axis .

b) at point p

field due to i2

B2 = Uo*i2/(2*pi*d) = 4*pi*10^-7*5.35/(2*pi*0.20) = 5.35*10^-6 T in -ve x-direction

so B1 = -5.35*10^-6 i

field due to i1

B1 = Uo*i1/(2*pi*1.414*d) = 4*pi*10^-7*3.10/(2*pi*1.414*0.20) = 2.192*10^-6 T ........135 degree from +ve x-axis

so B1 = (-1.54i + 1.54 j )*10^-6 T

NOW net magnetic field at p = - 6.89*10^-6 i + 1.54*10^-6 j

magnitude = [(-6.89)^2 + (1.54)^2]^0.5 = 7.05*10^-6 T

direction tan(theta) = 1.54/6.89 =0.223

theta = 12.57 degree

so direction is 167.42 degree from +ve x-axis

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