Physics - Magnetism Question with Solution | TestHub

Here, side of the triangle, $l=4.5\times 10^{-2}\mathrm{m}$, current, $\mathrm{I}=1\mathrm{A}$ magnetic field at the centre of the triangle ' $O$ ' $B=$ ? From figure, $\tan 60^{\circ }=\sqrt{3}=\frac{1}{2d}$ $$\Rightarrow d=\frac{l}{2\sqrt{3}}=\left(\frac{4.5\times 10^{-2}}{2\sqrt{3}}\right)\mathrm{m}$$
Magnetic field, $B=\frac{{\mu }_{0}i}{4\pi d}\left(\cos {\theta }_1+\cos {\theta }_2\right)$ Putting value of $\mu =4\pi \times 10^{-7}$ and ${\theta }_1$ and ${\theta }_2$ we will get net magenetic field $=3\times B=4\times 10^{-5}\mathrm{Wb}/{\mathrm{m}}^2$