Question 5 of 276

Two identical circular wires of radius $20cm$ and carrying current $\sqrt{2}A$ are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is________\times {10}^{-8}T.

(Take $\pi =3.14$)

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Answer: 628

The formula to calculate the magnetic field $\left(B\right)$ at the centre of a current carrying circular loop is given by
$B=\dfrac{{\mu }_{0}i}{2R}...(1)$
where, ${\mu }_{0}$ is the permeability of free space, $i$ is the current and $R$ is the radius of the loop.
${B}_{V},{B}_{H}$ are the magnetic field due to the vertical and the horizontal loop, which by magnitude are equal.
The resultant magnetic field $\left({B}_{net}\right)$ at the centre is, then, given by
${B}_{\text{net}}=\sqrt{{B}^{2}+{B}^{2}}=\sqrt{2}B =\dfrac{\sqrt{2}\times 4\pi \times {10}^{-7}\times \sqrt{2}}{2\times 0.2}$
$=628\times {10}^{-8}T$