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Solution,
Magnetic field due to a finite current carrying wire is given by the formula:
$B=\frac{{\mu }_{o}i}{4\pi a}\left(\mathrm{cos}\left({\theta }_{1}\right)+\mathrm{cos}\left({\theta }_{2}\right)\right)$

In the present case,
Thus, the magnetic field at the point P due to both the arms of the arrangement is
$B=2×\frac{{\mu }_{o}i}{4\pi a}\left(\mathrm{cos}\left({45}^{o}\right)+\mathrm{cos}\left({0}^{o}\right)\right)\phantom{\rule{0ex}{0ex}}B=2×\frac{{\mu }_{o}i}{4\pi a}\left(\frac{1}{\sqrt{2}}+1\right)\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{o}i\left(\sqrt{2}+1\right)}{2\sqrt{2}\pi a}$
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