Question 39 of 276

Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is $5.0cm$. The magnitude of the magnetic field at a point $P$ midway between the wires is _____\mu T. (Given: ${\mu }_{0}=4\pi \times {10}^{-7}Tm{A}^{-1}$)

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Answer: 160

The direction of the magnetic field at the observation point due to the current carrying conductor is determined by the right-hand cork screw rule.
From the given diagram, it is clear that as the conductors carry current in the mutually opposite direction, the direction of the magnetic fields produced by them at point $P$ will be the same.
Hence, the required magnetic field at $P$ due to the conductors can be calculated as follows:
$\begin{matrix}B & = & 2\times \left(\dfrac{{\mu }_{0}i}{2\pi a}\right) \\ & = & 2\times \dfrac{4\pi \times {10}^{-7}\times 10}{2\pi \times \left(\dfrac{5}{2}\times {10}^{-2}\right)}T \\ & = & 160\times {10}^{-6}T \\ & = & 160\mu T\end{matrix}$