Two long, parallel wires carry currents of I_1 = 3.22 A and I_2 = 4.55 A In the

Question

Two long, parallel wires carry currents of I_1 = 3.22 A and I_2 = 4.55 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 muT direction degree 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 4.55-A current. Magnitude muT direction degree to the left of the vertical

Explanation / Answer

a) at a point midway between the two wires,

Bnet = B2 - B1

= mue*I2/(2*pi*d/2) - mue*I1/(2*pi*d/2)

= (mue/pi*d)*(I2 - I1)

= (4*pi*10^-7/(pi*0.2))*(4.55 - 3.22)

= 2.66*10^-6 T

direction : 270 or 90 degrees with poitive x axis

b) at point P

B1 = mue*I1/(2*pi*d)

= 4*pi*10^-7*3.22/(2*pi*0.2)

= 3.22*10^-6 T

B1x = -3.22*10^-6 T

B1y = 0

B2 = mue*I2/(2*pi*d*sqrt(2))

= 4*pi*10^-7*3.22/(2*pi*0.2*sqrt(2))

= 2.28*10^-6 T

B2x = -B2*cos(45) = -1.61*10^-6 T

B2y = B2*sin(45) = 1.61*10^-6 T

Bx = B1x + B2x = -4.84*10^-6 T

By = B1y + B2y = 1.61*10^-6 T

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

= sqrt(4.84^2 + 1.61^2)*10^-6

= 5.1*10^-6 T

direction : theta = tan^-1(Bx/By)

= tan^-1(4.84/1.61)

= 71.6 degrees

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.