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  4. A conducting wire of uniform cross-section is bent to form an equilateral triangle ABC of side L . A current I enters at A and leaves at C . What will be the magnetic field at the centroid O of the triangle ?

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Q. A conducting wire of uniform cross-section is bent to form an equilateral triangle $ABC$ of side $L$ . A current $I$ enters at $A$ and leaves at $C$ . What will be the magnetic field at the centroid $O$ of the triangle ?

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
$B$ at $O:\left(i\right)\text{due to }I_{AB}\text{ is }B_{AB}=K\cdot \frac{I}{3}\bigodot$
$\left(\right.ii\left.\right)$ due to $I_{BC}$ is $B_{BC}=K\frac{I}{3}\bigodot$
$\left(\right.iii\left.\right)$ due to $I_{AC}$ is $B_{AC}=K\times \frac{2 I}{3}\bigotimes$
net $B$ at $O=K\frac{I}{3}+K\frac{I}{3}-K\frac{2 I}{3}=0$
here $K$ is any constant