Question

Problem 4. Two long, straight wires, each with a current of 12.0 A, are placed on two corners of an equilateral triangle with sides of length 2.50 cm as shown. Both of the wires have a current into the page.

What is the magnetic field at the third corner of the triangle? Sketch and label relevant vectors!

Another wire is placed at the third corner, parallel to the other two wires. Which direction (in or out of the page) should the current flow in the third wire so that the force on it is in the +y−direction. Explain briefly!

Answer 1

magnetic field due to current carrying wire = k * I / r

magnetic field due to wire 1 = 2 * 10^-7 * 12 / 0.025

magnetic field due to wire 1 = 0.000096 T

magnetic field due to wire 2 = 2 * 10^-7 * 12 / 0.025

magnetic field due to wire 2 = 0.000096 T

vertical component of the magnetic field will cancel each other as they are equal and in opposite direction

horizontal component of the fields = 0.000096 * cos(30) + 0.000096 * cos(30)

horizontal component of the fields = 0.00016627687 T

magnetic field at the third corner of the triangle = 0.00016627687 T

direction of the current in the third wire has to be out of the page so that force on it will be in +y direction

since the net magnetic field in + x direction, if the current is out of the page then by right hand rule the direction of force will be toward +y axis