There are two infinitely long straight current-carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductors is 1:1. The magnetic field at point P is:

We need to determine the correct expression for the total magnetic field at point $$P(x, y)$$ due to two semi-infinite straight conductors meeting at the origin $$O(0,0)$$ at right angles.

1. Calculate the Component Fields

From the layout shown , both wires carry an identical current $$I$$ and start from the origin, extending along the positive axes:

• Wire along the y-axis: It extends from $$(0,0)$$ to $$+\infty$$. The perpendicular distance to point $$P(x, y)$$ is $$x$$. By the right-hand rule, its magnetic field points into the page ($$-\hat{k}$$):

  $$B_y = \frac{\mu_0 I}{4\pi x} \left(1 - \frac{y}{\sqrt{x^2 + y^2}}\right)$$

• Wire along the x-axis: It extends from $$(0,0)$$ to $$+\infty$$. The perpendicular distance to point $$P(x, y)$$ is $$y$$. By the right-hand rule, its magnetic field points into of the page ($$+\hat{k}$$):

  $$B_x = \frac{\mu_0 I}{4\pi y} \left(1 - \frac{x}{\sqrt{x^2 + y^2}}\right)$$

2. Compute the Net Magnetic Field ($$B_{\text{net}}$$)

$$B_{\text{net}} = B_y + B_x = \frac{\mu_0 I}{4\pi x}\left(1 - \frac{y}{\sqrt{x^2 + y^2}}\right) + \frac{\mu_0 I}{4\pi y}\left(1 - \frac{x}{\sqrt{x^2 + y^2}}\right)$$

Factoring out the common constant $$\frac{\mu_0 I}{4\pi}$$ and combining the fractions over a common denominator ($$xy\sqrt{x^2 + y^2}$$) yields:

$$B_{\text{net}} = \frac{\mu_0 I}{4\pi xy\sqrt{x^2 + y^2}} \left[ y\sqrt{x^2 + y^2} - y^2 - x\sqrt{x^2 + y^2} + x^2 \right]$$

$$B_{\text{net}} = \frac{\mu_0 I}{4\pi xy\sqrt{x^2 + y^2}} \left[ (x^2 - y^2) + (x - y)\sqrt{x^2 + y^2} \right]$$

 Option C.