Physics - Magnetism Question with Solution | TestHub

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F = 2 ∫ dF sin θ

$=2\int \frac{{\mathit{\mu }}_0}{4\mathit{\pi }}\frac{2{\mathit{\text{j}}}_0\mathit{\text{dx}}\mathit{\text{.}}\mathit{\text{i}}}{\sqrt{{\mathit{\text{h}}}^2+{\mathit{\text{x}}}^2}}\mathit{\text{.}}\text{sin}\mathrm{\theta }$

$=\frac{{\mathit{\mu }}_0}{\mathit{\pi }}{\mathit{j}}_0\mathit{i}{{\displaystyle \int }}^{\mathit{\text{}}}\frac{\mathit{d}\mathit{x}}{\sqrt{{\mathit{h}}^2+{\mathit{x}}^{2^{}}}}\mathit{·}\frac{\mathit{h}}{\sqrt{{\mathit{h}}^2+{\mathit{x}}^2}}$

$=\frac{{\mathit{\mu }}_0}{\mathit{\pi }}{\mathit{\text{j}}}_0{\mathit{\text{i}}}_{·}\mathit{\text{h}}\underset{0}{\overset{\mathit{\text{d}}/2}{\int }}\frac{\mathit{\text{dx}}}{{\mathit{\text{h}}}^2+{\mathit{\text{x}}}^2}$

$=\frac{{\mathit{\mu }}_0}{\mathit{\pi }}{\mathit{\text{j}}}_0\mathit{\text{i}}\mathit{\text{h}}{\left[\frac{\mathit{\text{1}}}{\mathit{\text{h}}}{\text{tan}}^{-1}\frac{\mathit{\text{x}}}{\mathit{\text{h}}}\right]}_0^{\mathit{\text{d}}/2}$

$\overrightarrow{\mathit{\text{F}}}=\frac{{\mathit{\mu }}_0{\mathit{\text{j}}}_0\mathit{\text{i}}}{\mathit{\pi }}{\text{tan}}^{-1}\frac{\mathit{\text{d}}}{\mathit{\text{2h}}}\left(-\widehat{\mathit{\text{k}}}\right)$