Each of two long straight parallel wires 15 cm apart carries a current of 150 A.

Question

Each of two long straight parallel wires 15 cm apart carries a current of 150 A. The figure shows a cross section, with the wires running perpendicular to the page and point P lying on the perpendicular bisector of the line between the wires. Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is also out of the page. What is the magnitude of B? What is the value of Bx? What is the value of By? Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is into the page. What is the magnitude of B? What is the value of Bx? What is the value of By? What is the value of Bz?

Explanation / Answer

distance from wire to point P = r = (d/2)/sin45 = (0.15/2)/sin45 = 0.11 m

B1x = 0

B1y = -uo*I*cos45/(2*pi*r) = -(4*pi*10^-7*150*cos45)/(2*pi*0.11) = -0.0002 T

B1z = uo*I*sin45/(2*pi*r) = (4*pi*10^-7*150*sin45)/(2*pi*0.11) = 0.0002 T

B2x = 0

B2y = -uo*I*cos45/(2*pi*r) = -(4*pi*10^-7*150*cos45)/(2*pi*0.11) = -0.0002 T

B2z = -uo*I*sin45/(2*pi*r) = -(4*pi*10^-7*150*sin45)/(2*pi*0.11) = -0.0002 T

Bx = B1x + B2x = 0

By = B1y + B2y = -0.0004 T

Bz = B1z + B2z = 0

======================

when current in right wire is into the page

B1x = 0

B1y = -uo*I*cos45/(2*pi*r) = -(4*pi*10^-7*150*cos45)/(2*pi*0.11) = -0.0002 T

B1z = uo*I*sin45/(2*pi*r) = (4*pi*10^-7*150*sin45)/(2*pi*0.11) = 0.0002 T

B2x = 0

B2y = uo*I*cos45/(2*pi*r) = (4*pi*10^-7*150*cos45)/(2*pi*0.11) = 0.0002 T

B2z = uo*I*sin45/(2*pi*r) = (4*pi*10^-7*150*sin45)/(2*pi*0.11) = 0.0002 T

Bx = B1x + B2x = 0

By = B1y + B2y = 0

Bz = B1z + B2z = 0.0004 T

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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