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Question Number 81319 by rose last updated on 11/Feb/20

Answered by ajfour last updated on 12/Feb/20

Commented by ajfour last updated on 12/Feb/20

$${\overline{B}}_{AB}=\frac{{\mu }_0\left(2I/3\right)}{2\pi \left(\frac{L}{2\sqrt{3}}\right)}\left(\frac{\cos 30°+\cos 30°}{2}\right)$$$$=\frac{2{\mu }_{0}I}{2\pi L}$$$${\overline{B}}_{AC}+{\overline{B}}_{BC}=2\times \frac{{\mu }_0\left(I/3\right)}{2\pi \left(\frac{L}{2\sqrt{3}}\right)}\left(\frac{\cos 30°+\cos 30°}{2}\right)$$$$=\frac{{\mu }_{0}I}{2\pi L}\left(2\times \frac{1}{3}\times 2\sqrt{3}\times \frac{\sqrt{3}}{2}\right)◻$$$$=\frac{2{\mu }_{0}I}{2\pi L}◻$$$$hence$$$${\overline{B}}_{AB}+{\overline{B}}_{AC}+{\overline{B}}_{BC}=0.$$

Commented by ajfour last updated on 12/Feb/20

$$I/2indiagram,ishouldhave$$$$marked,I/3.$$

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