Ampere and Pythagoras

Let A be the point on the first wire closest to P, and let B be the point on the second wire closest to P. Clearly, the wires are perpendicular to the plane ABP and the triangle ABP is a right triangle and the right angle is at P, since 6 6+8 8=10*10. If we draw the magnetic field vectors generated by each of the conductors, these two vectors will be perpendicular to each other, no matter what the orientation of current flow is. The well-known formula for calculating the magnitude of the magnetic field near long thin wires is B = u0/2Pi * I/r, where u0 is the magnetic constant, I is the current and r is the distance from the wire. For the field of the first wire alone we get exactly 1/300000 teslas, and for the other one 1/400000 teslas. Since we know the vectors are perpendicular, we can get the resulting magnetic field strength by applying the Pythagoras' theorem: Br^2 = B1^2 + B2^2. From this we get Br = 1/240000 teslas, or 4.166667E-6

We start by noting that $6^{2}+8^{2}=10^{2}$ From Pythagorean Theorem, we know that the angle between the lines connecting point $P$ and the wires is $90^{\circ}$ . The total magnetic field at point $P$ is given by the vectoral sum, $\vec{B_{tot}}=\vec{B_1}+\vec{B_2}$ where $B_1$ and $B_2$ are the magnetic field inductions at point $P$ due to wire $1$ and wire $2$ respectively. From Ampere's Law we easily find that $B_k=\frac{\mu_0}{2\pi r_k}I_k$ for $k=1,2$ where $I_k$ $(k=1,2)$ are the currents in the wires $1$ and $2$ and $r_k$ $(k=1,2)$ are the distance from the point $P$ to the axes of the wires. Since $\vec{B_1}$ and $\vec{B_2}$ are pependicular to each other,the magnitude of the total magnetic field is equal to, $B_{tot}=\sqrt{B_1^{2}+B_2^{2}}$ $\Rightarrow B_{tot}=\frac{\mu_0}{2\pi}I\sqrt{\frac{1}{r_1^{2}}+\frac{1}{r_2^{2}}}$ $\Rightarrow B_{tot}=\frac{\mu_0}{2\pi} \times 1 \times \sqrt{\frac{1}{0.0036}+\frac{1}{0.0064}}$ $\Rightarrow B_{tot}=\boxed{4.2 \times 10^{-6}T}$

We know that the magnetic field created by a infinitely long, straight wire is $B = \frac {\mu_0 i} {2 \pi R}$ I will skip derivation of this formula for the sake of brevity. Then, we can find the magnitude of the magnetic field created by each wire by using the above formula. If we name two wires as A and B, then, $B_A = \frac {2 \times 10^{-7}} {.06} , B_B = \frac {2 \times 10^{-7}} {.08}$

Then, since the distance between two wires is 10cm, we can see that the axis point of two wires and point $P$ make a 3-4-5 triangle. That means that the two magnetic fields at point P are perpendicular to each other. Then, by using Pythagorean's Theorem, $B = \sqrt {B_A^2+B_B^2} = \boxed {4.1667 \times 10^{-7} T}$

Note that the direction of the current of the wire does not matter since in any case the two vectors will be perpendicular to each other.

Assume that the current flows in opposite directions (it hardly matters in which directions they flow. I've assumed it to make the image simple). image

Calculate the magnetic field due to each wire using $\frac{\mu_{o}}{2 \pi} \frac{i}{R}$ where $R$ is the perpendicular distance from the wire.

As you see, they are perpendicular in direction. So, the net magnetic field is $\frac{\mu_{o} i}{2 \pi} \sqrt{\frac{1}{R^{2}_1} + \frac{1}{R^{2}_2}}$ .

$R_{1}=0.06cm$ and $R_{2}=0.08cm$

The problem is quite intuitive and conceptual. Note that (6^2) + (8^2) = 100. Hence, now it becomes clear, that the given point is situated such that the lines joining the point and the axis of the wires, and the line joining the axes of 2 wires form a right angled triangle ( the line joining the axes of the wires being the hypotenuse). One more interesting and remarkable thing in this problem is the fact that the direction of current here is irrelevant. Just for verification, consider both the cases one by one. 1. First consider that current in both wires is in the same direction. Now find the direction of magnetic field at the point due to each wire. (note - To find the direction of magnetic field due to a long current carrying wire, keep your palm in the direction of current, and the curl your fingers towards the displacement vector joining the wire and the point. The direction of thumb is the direction of magnetic field).
So, you will notice that the magnetic fields due to the wires comes out to be mutually perpendicular.
2. Now consider the case that current in both wires is in opposite direction. Still, you will find that fields come out to be mutually perpendicular. So, the net magnetic field is the vector addition of both mutually perpendicular magnetic fields.

To calculate the field due to a long wire at a point, use the following formula