Question

Three infinitely long parallel wires carrying equal currents of 2.5 A are placed at each corner of an equilateral triangle whose side is 3.0 cm. Two currents are directed into the page and the third current is out of the page.

a)(5 points) What is the magnitude and direction of the magnetic field at the center of the triangle?

b)(5 points) What is the magnitude and direction of the magnetic force on a 1.5 A current in a 2.0 m wire placed at the center of the triangle parallel to the other three wires and directed into the page?

I have an answer key that says:

From symmetries and that the B fields from the two lower currents from two side of an equilateral triangle whose third side is the vector sum, these fields add to a single B of the same value which is directed horizontally to the right. This is collinear to the B field of the third charge. They add to give the total B field B = 2Bi = 2(2.88 x 10– 5 T ) = 5.76 x 10– 5 T = 5.8 x 10– 5 T.

Which I really just do not understand, why do the two fields directed inward add to a single B of the same value. when I am solving i get that they both have a value of 2.88cos(45)=2.04 to the right, so wouldn't they have a vector addition of 4.08 not 2.88? Can someone give a more in depth explanation?

Answer 1

Solution:

Part (a): Let one side of the equilateral tringle is along the X- axis.

The magnitude of magnetic field at the centre of equilatral tringle due to a single wire is given by

$B=\frac{\mu _{0}I}{2\pi d}\\\\ \Rightarrow B=\frac{\mu _{0}}{4\pi }\times \frac{2I}{d}\\\\ \left ( \because \frac{\mu _{0}}{4\pi }=10^{-7}TmA^{-1}, I=2.5A \ and \ d= \sqrt{3} cm=13.32\times 10^{-3}m \right )\\\\ \Rightarrow B=\frac{10^{-7}\times 2\times 2.5}{17.32\times 10^{-3}}\\\\ \Rightarrow B=2.887\times 10^{-5}T$ (1)

Since, the angle between any two vertex making at the centre of equilateral tringel is $120^{\circ}$ . If all the currents are flowing in the same direction, then the magnetic field produced by these current at the centre of equilateral tringle also making an angle $120^{\circ}$ from each other and the resultant magnetic field at the centre is equal to zero by the triangle rule of vector sum. But here two currents flowing into the page and one out of the page. So, the resultant of two current flowing into the page and one out the page are paraller and same in magnitude. Thus the magnutude of resultant magnetic field at the centre is given by

$B_{centre}=2B\\\\ \Rightarrow B_{centre}=2\times 2.887\times 10^{-5}\\\\ \Rightarrow B_{centre}=5.774\times 10^{-5}T\\\\$ (2)

and its direction is $140^{\circ}$ from the positive X- axis.

Part (b): The magnitude of magnetic force at the centre of triangle is given by

$F=IB_{centre}\int dl\\\\ \Rightarrow F=IB_{centre}L\\\\ \left ( \because B_{centre}=5.774\times 10^{-5}T, I=1.5A \and \ L=2.0m \right )\\\\ \Rightarrow F=1.5\times 5.774\times 10^{-5}\times 2.0\\\\ \Rightarrow F=1.732\times 10^{-4}N$ (3)

And the direction of force is perpendicular to the plane containing B_{centre} and current I =1.5 A.