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  4. A straight section PQ of a circuit lies along x-axis at x=-(a/2) to x=+(a/2) and carries a steady current (i), the magnetic field due to section PQ at a point x = + a will be

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Q. A straight section PQ of a circuit lies along x-axis at $ x=-\frac{a}{2}\,\,\,to\,\,\,x=+\frac{a}{2} $ and carries a steady current (i), the magnetic field due to section PQ at a point x = + a will be

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Solution:

From Biot Savarts law, magnetic field at point R due to a straight wire $ B=\frac{{{\mu }_{0}}}{4\pi }\frac{i\Delta l\sin \theta }{{{r}^{2}}} $ but $ \theta =0 $ $ \therefore $ $ B=0 $