Three very long,straight,parallel wires each carry currents of 4.00 A,directed out of the page as shown in the figure.The wires pass through the vertices of a right isosceles triangle of side 2.00 cm.What is the magnitude of the magnetic field at point P at the midpoint of the hypotenuse of the triangle? A)4.42 × 10^-6 T B)1.77 × 10^-5 T C)5.66 × 10^-5 T D)1.26 × 10^-4 T E)1.77 × 10^-6 T

Question

Quizplus · Accepted Answer

The magnetic field at point P due to each wire can be determined using the formula: 
$B=\frac{\mu_0}{4\pi}\frac{I}{r}$ 
where $\mu_0$ is the permeability of free space, $I$ is the current in the wire, and $r$ is the distance from the wire to the point of interest. 
The magnetic fields due to the top and bottom wires will be equal in magnitude and cancel out at point P because they are directed in opposite directions. The magnetic field due to the middle wire can be found by treating it as a straight wire carrying a current of 8.00 A. The distance from the wire to point P is equal to the length of the hypotenuse of the right isosceles triangle, which is $2\sqrt{2}$ cm. Thus: 
$B=\frac{\mu_0}{4\pi}\frac{8.00	ext{ A}}{2\sqrt{2}	ext{ cm}}= 5.66 	imes 10^{-5} 	ext{ T}$ 
Therefore, the total magnetic field at point P is equal to the magnitude of the magnetic field due to the middle wire: 
$B_{total}=B=5.66 	imes 10^{-5} 	ext{ T}$ 
The correct answer is C.

Three Very Long,straight,parallel Wires Each Carry Currents of 4