Question

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 conductor is 1 : 1. The magnetic field at point P is ____________.

• A. \(\frac{\mu_0 I}{4\pi xy}\left[\sqrt{x^2+y^2} + (x+y)\right]\)
• B. \(\frac{\mu_0 I}{4\pi xy}\left[\sqrt{x^2+y^2} - (x+y)\right]\)
• C. \(\frac{\mu_0 I xy}{4\pi}\left[\sqrt{x^2+y^2} - (x+y)\right]\)
• D. \(\frac{\mu_0 I xy}{4\pi}\left[\sqrt{x^2+y^2} + (x+y)\right]\)

Hint:

Use the semi-infinite wire result \(B = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)\) for each wire and check the directions with the right-hand rule: both fields at P point the same way, so they add.

Answer 1

The Correct Option is A

Let \(r = \sqrt{x^2+y^2}\). Wire along the y-axis (current upward, from O to infinity) is at perpendicular distance x from P. Its ends subtend angles with the perpendicular from P: the far end gives \(\sin\alpha = 1\), and the end O gives \(\sin\alpha = \frac{y}{r}\) (on the same side as the foot of the perpendicular, so it adds).
\(B_1 = \frac{\mu_0 I}{4\pi x}\left(1 + \frac{y}{r}\right)\).
Wire along the x-axis (current towards O, from infinity) is at perpendicular distance y from P: \(B_2 = \frac{\mu_0 I}{4\pi y}\left(1 + \frac{x}{r}\right)\).
By the right-hand rule, both fields at P are directed into the plane, so they add:
\(B = \frac{\mu_0 I}{4\pi}\left[\frac{1}{x}+\frac{1}{y}+\frac{y}{xr}+\frac{x}{yr}\right] = \frac{\mu_0 I}{4\pi}\left[\frac{x+y}{xy} + \frac{x^2+y^2}{xyr}\right]\)
\(= \frac{\mu_0 I}{4\pi xy}\left[(x+y) + \sqrt{x^2+y^2}\right]\).
Answer: (1) \(\frac{\mu_0 I}{4\pi xy}\left[\sqrt{x^2+y^2} + (x+y)\right]\)