Consider two current carrying wires shown below. The current in the wire 1 is 4.5 A in the positive x direction. Find the magnitude and direction of the current in wire 2 if the net magnetic field at point A is to be 7.5 μT in the positive z direction (coming out of the paper).

additional information:the distance between wire 1 and wire is 25 cm and the distance between the wire 2 and point A is 12 cm.

To find the magnitude and direction of the current in wire 2, we can use the Biot-Savart law, which relates the magnetic field produced by a current-carrying wire to the current and the distance from the wire.

The Biot-Savart law states that the magnetic field (\( \mathbf{B} \)) produced by a current (\( \mathbf{I} \)) flowing through a wire at a distance (\( \mathbf{r} \)) from the wire is given by:

\[ d\mathbf{B} = \frac{{\mu_{0}}}{4\pi} \frac{{\mathbf{I}\times \mathbf{r}}}{r^3} \]

where \( \mu_0 \) is the permeability constant.

To find the magnitude and direction of the current in wire 2, we need to find the distance between the two wires at point A, and set up an equation using the Biot-Savart law with the known values.

Let's assume the current in wire 2 is \( I_2 \). The distance (\( r \)) between the two wires at point A is the perpendicular distance from wire 1 to wire 2.

To find the perpendicular distance between two parallel wires, we can use the concept of the right-hand rule. If we let wire 1 be the thumb and wire 2 be the index finger of our right hand, then the perpendicular distance is the distance between the two fingers.

Given that the net magnetic field at point A is 7.5 μT (coming out of the paper), we can set up the following equation:

\( \mathbf{B_{net}} = \frac{{\mu_{0}}}{4\pi} \frac{{\mathbf{I_1}\times \mathbf{r_1}}}{r_1^3} + \frac{{\mu_{0}}}{4\pi} \frac{{\mathbf{I_2}\times \mathbf{r_2}}}{r_2^3} \)

Here, \( \mathbf{I_1} \) is the current in wire 1, \( \mathbf{r_1} \) is the distance vector from wire 1 to point A, \( \mathbf{I_2} \) is the current in wire 2, \( \mathbf{r_2} \) is the distance vector from wire 2 to point A, and \( \mu_0 \) is the permeability constant.

To find the direction of the current in wire 2, we need to determine the direction of the magnetic field produced by wire 2 at point A. The direction of the magnetic field can be obtained using the right-hand rule, where the curled fingers of the right hand point in the direction of the current, and the extended thumb points in the direction of the magnetic field.

Solving the equation using the known values, we can find the magnitude and direction of the current in wire 2.