Question
The. wire ABC shown in figure forms an equilateral triangle. Find the magnetic field B at the centre O of the triangle assuming the wire to be uniform.
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Answer
For AB B is along $\odot\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ+\sin60^\circ)$For AC B $\otimes\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ+\sin60^\circ)$
For BD B $\odot\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ)$
For DC $\otimes\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ)$
$\therefore\ \text{Net}\ \text{B}=0$
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