Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (See figure) ( \mu 0 = 4 \pi × 10–7 N-A–2)

Magnetics PYQ — JEE Main 2019 · 12 Apr · Shift 2 · Q66 with Solution

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Correct answer: A

Magnetic field due to a finite straight current-carrying wire

The magnetic field at a point at perpendicular distance $r$ from a finite wire is
$B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)$
where $\theta_1$ and $\theta_2$ are the angles subtended by the two ends of the wire at the point.
From the figure

Since the wire segment $AB$ has length $6\text{ cm}$ and point $P$ is opposite its midpoint at a perpendicular distance $4\text{ cm}$ , the half-length is
$\frac{AB}{2} = 3\text{ cm}$
So from geometry,
$\sin\theta = \frac{3}{\sqrt{3^2+4^2}} = \frac{3}{5}$
Thus,
$\theta_1 = \theta_2 = \theta, \quad \sin\theta_1 + \sin\theta_2 = 2\times \frac{3}{5} = \frac{6}{5}$
Substitute values

Given:
$\mu_0 = 4\pi \times 10^{-7}\,\text{N A}^{-2}, \quad I=5\,\text{A}, \quad r=4\text{ cm}=0.04\,\text{m}$
Therefore,
$B = \frac{4\pi\times 10^{-7}\times 5}{4\pi\times 0.04}\times \frac{6}{5}$
Simplifying,
$B = \frac{10^{-7}\times 5}{0.04}\times \frac{6}{5}$ $B = \frac{10^{-7}\times 6}{0.04}$ $B = 1.5\times 10^{-5}\,\text{T}$
Match with options
$\boxed{1.5\times 10^{-5}\,\text{T}}$
So the correct option is A.