JEE MainPhysicsModerateMagnetic effect of current

The magnitude of magnetic field at O (centre of the circular
part) of the current carrying coil as shownis

Show answer and solution

Magnetic field due to circular
segment

= \frac{3}{4} \cdot \frac{\mu_{0}i}{2a}

Due to one straight wire
segment

= \frac{\mu_{0}i}{4\pi b}\left( sin45^{\circ} + sin0^{\circ} \right) = \frac{\mu_{0}i}{4\sqrt{2}\pi b}

Net field

= \frac{3\mu_{0}i}{8a} + 2 \times \frac{\mu_{0}i}{4\sqrt{2}\pi b} = \frac{\mu_{0}i}{4\pi}\left( \frac{3\pi}{2a} + \frac{\sqrt{2}}{b} \right)

Source: Grass-G / P-13 Physics paper 4 Nov.docx

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