There are two infinitely long straight current carrying conductors… | JEE Main 2021 (1 Sep, Shift 2)

Q51JEE Main 2021 · 1 SEP · SHIFT 2Medium

Magnetic Field of Straight Wires

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 ____________.

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Answer: (a)

Finite-wire formula: $B = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$ .
Wire on x-axis ( $d = y$ ): $B_1 = \frac{\mu_0 I}{4\pi y}\left(1 + \frac{x}{\sqrt{x^2+y^2}}\right)$ ; wire on y-axis ( $d = x$ ): $B_2 = \frac{\mu_0 I}{4\pi x}\left(1 + \frac{y}{\sqrt{x^2+y^2}}\right)$ .
Both point into the page, so they add: $B = \frac{\mu_0 I}{4\pi xy}\left[x + y + \frac{x^2+y^2}{\sqrt{x^2+y^2}}\right]$
$B = \frac{\mu_0 I}{4\pi xy}\left[\sqrt{x^2+y^2} + (x+y)\right]$
So the answer is (a).

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Magnetic Field of Straight Wires question from JEE Main 2021 (1 Sep, Shift 2)

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