The magnitude of magnetic induction at the mi | vedclass

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(D) The arrangement consists of two semi-infinite wires and two quarter-infinite segments. However, looking at the geometry, the magnetic field at poi

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$\frac{\mu_0 I}{\pi a}$

Vedclass · Answer

(D) The arrangement consists of two semi-infinite wires and two quarter-infinite segments. However, looking at the geometry, the magnetic field at point '$O$' due to the segments $AB$ and $BC$ can be calculated using the Biot-Savart Law. For a semi-infinite wire, the magnetic field at a perpendicular distance $a$ is $B = \frac{\mu_0 I}{4\pi a}$. In this configuration, the segments $AB$ and $BC$ create a magnetic field at '$O$' directed into the page. Similarly, the segments $TE$ and the other horizontal segment create a magnetic field at '$O$' also directed into the page. By symmetry and applying the right-hand rule, the contributions from the four segments add up. Specifically, each segment acts as a semi-infinite wire relative to point '$O$'. The total magnetic field $B_{total} = 4 	imes (\frac{\mu_0 I}{4\pi a}) = \frac{\mu_0 I}{\pi a}$.

The magnitude of magnetic induction at the midpoint '$O$' due to the current arrangement shown in the figure is ($\mu_0$ = permeability of free space).

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