Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is 5.0 ~cm . The magnitude of the magnetic field at a point P midway between the wires is \underline\hspace2cm \mu T (Given : \mu_0=4 \pi \times 10^-7 TmA^-1 )

Magnetics PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q88 with Solution

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Correct answer: 160

Magnetic field due to one long straight wire

For a long straight wire carrying current $I$ , the magnetic field at distance $r$ is
$B=\frac{\mu_0 I}{2\pi r}$
Distance of point $P$ from each wire

The wires are separated by $5.0\text{ cm}$ , and $P$ is midway between them.

Hence,
$r=\frac{5.0}{2}\text{ cm}=2.5\text{ cm}=2.5\times 10^{-2}\text{ m}$
Direction of magnetic fields at midpoint

Since the currents are equal and in opposite directions, the magnetic fields at the midpoint are in the same direction.

Therefore, they add.
Field due to one wire

From the figure, each wire carries current $I=10\text{ A}$ .

So,
$B_1=\frac{\mu_0 I}{2\pi r} =\frac{4\pi\times 10^{-7}\times 10}{2\pi\times 2.5\times 10^{-2}}$
Simplifying,
$B_1=\frac{4\times 10^{-6}}{5\times 10^{-2}}=8\times 10^{-5}\text{ T}$
Converting to microtesla,
$8\times 10^{-5}\text{ T}=80\,\mu\text{T}$
Net magnetic field at midpoint

Since both fields add,
$B_{\text{net}}=2B_1=2\times 80=160\,\mu\text{T}$
Final answer
$\boxed{160}$