Two identical circular wires of radius 20 cm and carrying current 2 A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is ________ × 10

Question

Two identical circular wires of radius 20 cm and carrying current 2 A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is ________ × 10 - 8 T . (Take π = 3 . 14 )

TestHub · Accepted Answer

The formula to calculate the magnetic field $(B)$ at the centre of a current carrying circular loop is given by
$B = \frac{μ_{0} i}{2 R} . . . (1)$
where, $μ_{0}$ 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 $(B_{n e t})$ at the centre is, then, given by
$B_{net} = \sqrt{B^{2} + B^{2}} = \sqrt{2} B = \frac{\sqrt{2} \times 4 π \times 10^{- 7} \times \sqrt{2}}{2 \times 0.2}$
$= 628 \times 10^{- 8} T$

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