Physics - Magnetism Question with Solution | TestHub

Due to current through side AB Magnetic field at the centre $\mathrm{O}$ ${\mathrm{B}}_1=\frac{{\mu }_0\mathrm{I}}{4\pi \mathrm{a}}\left[\sin {\theta }_1+\sin {\theta }_2\right]$
As the magnetic field due to each of the three sides is the same in magnitude and direction. $\therefore$ Total magnetic field at $\mathrm{O}$ is sum of all the fields. i.e. $\mathrm{B}=3{\mathrm{B}}_1=\frac{3{\mu }_0\mathrm{I}}{4\pi \mathrm{a}}\left[\sin {\theta }_1+\sin {\theta }_2\right]$ Here, $\tan {\theta }_1=\frac{\mathrm{AD}}{\mathrm{OD}}\Rightarrow \tan 60^{\circ }=\frac{\frac{\ell }{2}}{\mathrm{a}}$ $\Rightarrow a=\frac{\ell }{2\sqrt{3}}=\frac{9\times 10^{-2}}{2\sqrt{3}}$ Now B $\begin{array}{rl}& =3\times \frac{4\pi \times 10^{-7}\times 2}{4\pi \times \frac{9\times 10^{-2}}{2\sqrt{3}}}\left[\sin 60^{\circ }+\sin 60^{\circ }\right]\\ =& \frac{4\sqrt{3}}{9}\times 10^{-5}\left[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]\\ =& 1.33\times 10^{-5}\mathrm{T}\end{array}$