Question

# Write Ampere's circuital law. A long straight wire of a circular cross-section (radius $$a$$) carrying steady current. Current is uniformly distributed in the wire. Calculate magnetic field inside the region $$(r< a)$$ in the wire.

Solution

## According to Ampere's circuital law, integral of magnetic field along a closed curve equals the sum of all electric currents passing through the cross section of the closed curve times permeability.$$\oint \vec I. \vec {dl} = \mu_o I_{enclosed}$$To find magnetic field inside a long straight wire of cross-sectional area $$a$$, refer the attached figure. Magnetic field lines in this configuration forms concentric circles centered at $$O$$, the axis of the cylindrical wire. Hence, it is the preferred path of the Amperian Loop.Consider an Amperian Loop of radius $$a<r$$ as shown in the figure.Current enclosed in a cross-section area $$\pi a^2$$ is $$I$$Then, Current enclosed in a cross-section area $$\pi r^2$$ is $$I_{enclosed} = \cfrac{r^2}{a^2}I$$Taking the magnetic field integral along the loop, we get$$\oint \vec B \cdot \vec{dl} = B \times 2 \pi r$$Using Ampere's Law, we get$$B \times 2 \pi r = \cfrac{ \mu_o I r^2}{a^2}$$$$B = \cfrac{\mu_o r I}{2 \pi a^2}$$Direction of magnetic field can be found using Fleming's Right hand thumb rule.Physics

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