Answer:
For the current carrying loop quarter circles of radius R, lying in the positive quadrants of the XY-plane. \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{I(\pi /2)}{R}\hat{k}=\frac{{{\mu }_{0}}}{4}\frac{I}{2R}\hat{k}\] For the current carrying loop quarter circles of radius R, tying in the positive quadrants of the YZ-plane magnetic field at the centre. \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4}\frac{I}{2R}\hat{i}\] Similarly, for the current carrying loop quarter circles of radius R, lying in the positive quadrants of the ZX-plane \[{{B}_{3}}=\frac{{{\mu }_{0}}}{4}\frac{I}{2R}\hat{j}\] Current carrying loop consists of three identical quarter circles of radius R, lying in the positive quadrants of the X-Y, V-Z and Z-X planes with their centres at the origin, joined together is equal to the vector sum of magnetic field due to each quarter and given by \[B=\frac{1}{4\pi }(\hat{i}+\hat{j}+\hat{k})\frac{{{\mu }_{0}}I}{2R}\]
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