Four very long, current carrying wires in the same plane intersect to form a square 40.0 cm on each side, as shown in Fig E 28.26. Find the magnitude and direction of the current so that the magnetic field at the center of the square is zero.

The magnetic field due to wire at the point is given by:

$B=\frac{{\mu }_{0}I}{2\pi r}$

Where B is the magnetic field due to wire,${\mu }_0$ is the permeability of the vaccum, l is the current through the wire, and r is the distance from the wire.

Sign convention for magnetic field:

Conventionally for, the direction of the inward and outward magnetic field to the plane negative or positive sign is considered for the magnetic field.

Given data:

• The magnetic field at the center is zero.
• The current flow in wire 1, wire 2, and wire 3are $l_1=10A,l_2=8A,$and$l_3=20A$ respectively.

The magnetic field due to wire 1 at the center is given by:

$B_1=-\frac{{\mu }_0{l}_1}{2\pi r}$

Here, the Magnetic field is negative due to the inward direction of the magnetic field with the plane.

The magnetic field due to wire 2 at the center is given by:

$B_2=-\frac{{\mu }_0{I}_2}{2\pi r}$

Here, the Magnetic field is negative due to the inward direction of the magnetic field with the plane.

The magnetic field due to wire ₃ at the center is given by:

$B_3=\frac{{\mu }_0{I}_3}{2\pi r}$

Here, the Magnetic field is positive due to the outward direction of the magnetic field with the plane.

The magnetic field due to wire 4 at the center is given by:

$B_4=\frac{{\mu }_0{I}_4}{2\pi r}$

Here, the Magnetic field is positive due to the outward direction of the magnetic field with the plane.

The current in wires flows in such a manner that causes the net magnetic field at the center of the square to become zero.

So,

$\begin{array}{r}{B}_1+B_2+B_3=B_4\\ -\frac{{\mu }_0{I}_1}{2\pi r}-\frac{{\mu }_0{I}_2}{2\pi r}+\frac{{\mu }_0{I}_3}{2\pi r}=\frac{{\mu }_0{I}_4}{2\pi r}\\ -I_1-I_2+I_3=I_4\end{array}$

Now, putting the values of constants in the above equation, and we get,

$\begin{array}{r}-10\mathrm{A}-8\mathrm{A}+20\mathrm{A}=I_4\\ I_4=2.0\mathrm{A}\end{array}$

Thus, the magnitude and direction of current in the wire are and towards the bottom of the plane, respectively.