Question 51 of 276

Two long parallel wires carrying currents $8A$ and $15A$ in opposite directions are placed at a distance of $7cm$ from each other. A point $P$ is at equidistant from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnitude of magnetic field at $P$ is ______ $\times {10}^{-6}T$. (Given: $\sqrt{2}=1.4$)

Your answerCheck answer

Show answer and solution▾

Answer: 68

Here, the magnetic fields produced by the two wires will be perpendicular to one another.
Magnetic field due to wire carrying current $8A$ is ${B}_{1}=\dfrac{{\mu }_{0}{i}_{1}}{2\pi d}$
and magnetic field due to wire carrying current $15A$ is ${B}_{2}=\dfrac{{\mu }_{0}{i}_{2}}{2\pi d}$
The net magnetic field is
${B}_{net}=\sqrt{{B}_{1}^{2}+{B}_{2}^{2}}=\dfrac{{\mu }_{0}}{2\pi d}\sqrt{{i}_{1}^{2}+{i}_{2}^{2}}$
$=\dfrac{4\pi \times {10}^{-7}}{2\pi \times \left(\dfrac{7}{\sqrt{2}}\right)\times {10}^{-2}}\times \sqrt{{8}^{2}+{15}^{2}}\left(d=\dfrac{7}{\sqrt{2}}cm\right)$
$=68\times {10}^{-6}T$.