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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.

631433_604122_ans_6ebe8b1217ee428ab1b97ebb47b0ec70.png

Physics

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