A very long wire ABDMNDC is shown in the figu | vedclass

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Question

(B) The total magnetic field $B$ at the center is the vector sum of the fields produced by the three parts: the straight wire $AB$,the circular loop $

Vedclass · Accepted Answer

$\frac{\mu_{0} I}{2 \pi R}(\pi+1)$

Vedclass · Answer

(B) The total magnetic field $B$ at the center is the vector sum of the fields produced by the three parts: the straight wire $AB$,the circular loop $DMND$,and the straight wire $BC$.$1$. The magnetic field due to the straight wire $AB$ at the center of the circle (which is at a perpendicular distance $R$ from the line of $AB$): $B_{AB} = \frac{\mu_{0} I}{4 \pi R} (\sin 90^{\circ} + \sin 0^{\circ}) = \frac{\mu_{0} I}{4 \pi R}$.$2$. The magnetic field due to the circular loop $DMND$ at its center: $B_{loop} = \frac{\mu_{0} I}{2 R}$.$3$. The magnetic field due to the straight wire $BC$ at the center of the circle (which is at a perpendicular distance $R$ from the line of $BC$): $B_{BC} = \frac{\mu_{0} I}{4 \pi R} (\sin 90^{\circ} + \sin 0^{\circ}) = \frac{\mu_{0} I}{4 \pi R}$.Since the current flows in the same sense for all parts,the magnetic fields add up:$B = B_{AB} + B_{loop} + B_{BC} = \frac{\mu_{0} I}{4 \pi R} + \frac{\mu_{0} I}{2 R} + \frac{\mu_{0} I}{4 \pi R}$$B = \frac{\mu_{0} I}{2 \pi R} + \frac{\mu_{0} I}{2 R} = \frac{\mu_{0} I}{2 \pi R} (1 + \pi)$.

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