Physics - Magnetism Question with Solution | TestHub

The magnetic field in between because of each will be in opposite direction

$B_{\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}}=\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi }x} \hat{\mathbf{j}}-\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi } (2d-x)} (-\hat{\mathbf{j}})$

$=\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi }}\left[\frac{1}{x}-\frac{1}{2d-x}\right] \left(\hat{\mathbf{j}}\right)$

At$x=d, B_{\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}}=0$

For$x<d, B_{\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{ }}=\left(\widehat{\mathbf{j}}\right)$

For$x>d, B_{\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}}=(-\widehat{\mathbf{j}})$

Towards$x,$net magnetic field will add up and direction will be$\left(-\widehat{\mathbf{j}}\right).$

Towards$x^{\prime },$net magnetic field will add up and direction will be$\left(-\widehat{\mathbf{j}}\right).$