As shown in the figure below, two long parallel wires (1 and 2) carry currents o

Question

As shown in the figure below, two long parallel wires (1 and 2) carry currents of I1 = 2.90 A and I2 = 4.90 A in the direction indicated.

(a) Determine the magnitude and direction of the magnetic field at a point midway between the wires (d = 10.0 cm).

(b) Determine the magnitude and direction of the magnetic field at point P, located d = 10.0 cm above wire 1.

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 = 2.90 A and I2 = 4.90 A in the direction indicated. (a) Determine the magnitude and direction of the magnetic field at a point midway between the wires (d = 10.0 cm). (b) Determine the magnitude and direction of the magnetic field at point P, located d = 10.0 cm above wire 1.

Explanation / Answer

r = d/2 = 5 cm = 0.05 m

a) B1 = uo*I1/(2*pi*r) = (4*3.14e-7*

B2 = uo*I2/(2*pi*r)

at mid point B1 &B2 are in opposite direction

Bnet = B1 - B2 = uo*(I1-I2)/*2*pi*r)

Bnet = (4*3.14e-7*(2.9-4.9))/(2*3.14*0.05) = -80 uT

direction 90 counterclockwise from the +x-axis

b) r = sqrt(d^2 + d^2) = sqrt2*d = 14.14 cm = 0.141 m

B1x = uo*I1/2*pi*d = 5.8e-6

B1y = 0

B2x = uo*I2*cos45/(2*pi*r) = (2e-7*4.9*cos45)/(0.141) = 4.9e-6 T

B2y = uo*I2*sin45/(2*pi*r) = 4.9e-6 T

Bx = B1x + B2x = 10.7 uT

By = B1y + B2y = 4.9 uT

Bnet = sqrt(Bx^2+By^2) = 11.77 uT

direction = tan^-1(By/Bx) = 24.6

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