Each of two long straight parallel wires 12 cm apart carries a current of 150 A

Question

Each of two long straight parallel wires 12 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.

A. What is the magnitude of B?

B. What is the value of Bx?

C. What is the value of By?

D. What is the value of Bz?

E. 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?

F. What is the value of Bx?

G. What is the value of By?

Explanation / Answer

Magnetic field strength can be found using the formula:

Force/length = k(Current1xCurrent2/distance)

Which can be rearranged to:

F = k(current1xcurrent2/distance)xlength

k (or the magnetic force constant) = 2.0x10^-7

current in wire1 =150A

crrent in wire2 =150A

distance =12cm or 0.12m

F=2x10-7 ((150x150)/ 0.12)

F= 0.0187 or 18x10-3

However, we are not finished yet. Since force is a vector quantity, we also need to give a direction. Since the opposite parallel wires have currents travelling in opposite directions, the force attract them.

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