91Ó°ÊÓ

We need to find the direction of both magnitude fields.

The current in both wire is in the same direction with the same magnitude. a megnatic field produce by the current-carrying wires and the Biot-Savart law enables us to calculate the magnitude and direction of this magnetic field. at any point where the magnetic field due to the segment $△\stackrel{→}{s}$of current-carrying wire is given by equation

$B=\frac{{\mathrm{μ}}_{o}I}{2\mathrm{Ï€°ù}}\left(1\right)$

The distance between the wire and the point known as $r$and The current of the wire known as $I$At point $\left(1\right)$, $4cm$is the distance between it and the intersection . So, its distance from the horizontal wire is

$r_1=r_2=3.86cm$

From the second wire the same distance are given

$r_1=r_2=3.86cm$

After applying right hand rule, Each wire has direction out of the page after finding the magnetic field at point 1, so the magnetic field at point 1 is the summation of both magnetic fields

$B_1=\frac{{\mathrm{μ}}_o}{2\mathrm{π}}\frac{I}{r_1}+\frac{{\mathrm{μ}}_o}{2\mathrm{π}}\frac{I}{r_2}=2\left(\frac{{\mathrm{μ}}_o}{2\mathrm{π}}\frac{I}{r_1}\right)\left(2\right)$

The values for and to get the magnetic field at point 1 by

$$\begin{gathered} B_1=2\left(\frac{{\mathrm{μ}}_o}{2\mathrm{π}}\frac{I}{r_1}\right) \\ =2\left(\frac{\left((4\mathrm{π}×{10}^{-7}\mathrm{T})×mA\right)\left(5A\right)}{2\mathrm{π}\left(3.86×{10}^{-2}\mathrm{m}\right)}\right) \\ =5.2×{10}^{-5}T \end{gathered}$$

By using right-hand rule, you determine the direction of the fields by pointing your right thumb in the direction of current while the remaining fingers curl around the wire and point in the direction of magnetic field.

A magnetic field is applied at point $2$by the horizontal wire with its direction out of the page while the other wire has its direction in. In the case where both wires exert the same magnetic field at point $2$, then the net magnetic field at point $2$is zero

$B_2=0$