1 and I2 respectively. What is the magnetic field at a point (2a,2a) on the z-axis?

Assuming that the currents are flowing in the positive direction of the axes, the magnetic field due to the wire carrying current I1 along the x-axis at point (2a,2a) on the z-axis is given by:

B1 = (μ0/4π) * I1 * ∫ dl1 × r1 / r1^3

where μ0 is the permeability of free space, dl1 is an element of length along the wire carrying current I1, r1 is the vector joining the element dl1 to the point (2a,2a) on the z-axis, and the integral is taken along the entire length of the wire.

Similarly, the magnetic field due to the wire carrying current I2 along the y-axis at point (2a,2a) on the z-axis is given by:

B2 = (μ0/4π) * I2 * ∫ dl2 × r2 / r2^3

where dl2 is an element of length along the wire carrying current I2, r2 is the vector joining the element dl2 to the point (2a,2a) on the z-axis, and the integral is taken along the entire length of the wire.

The total magnetic field at point (2a,2a) on the z-axis due to both wires is the vector sum of B1 and B2:

B = B1 + B2

This calculation can be simplified by noting that the vectors r1 and r2 have the same magnitude and are at an angle of 45° to the x and y axes. Therefore, we can write:

r1 = r2 = a√2 î + a√2 ĵ + 2a k̂

where î, ĵ, and k̂ are unit vectors along the x, y, and z axes respectively.

Using this expression for r1 and r2, we can evaluate the integrals for B1 and B2, and obtain:

B1 = (μ0/4π) * I1 * (a√2 ĵ - a√2 î) / 2a^2

B2 = (μ0/4π) * I2 * (-a√2 î - a√2 ĵ) / 2a^2

Substituting these expressions into the equation for B, we get:

B = (μ0/4π) * [(I1 - I2) a√2 î + (I2 - I1) a√2 ĵ] / 2a^2

Simplifying, we obtain:

B = (μ0/4π) * (I1 - I2) √2 k̂ / 2a

Therefore, the magnetic field at point (2a,2a) on the z-axis due to the two wires carrying currents I1 and I2 is:

B = (μ0/4π) * (I1 - I2) √2 / (2a) k̂