Question
A long cylindrical wire of radius b carries a current i distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point inside the wire at a distance a from the axis.
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Answer
i is uniformly distributed throughout So, ‘i’ for the part of radius $\text{a}=\frac{\text{i}}{\pi\text{b}^2}\times\pi\text{a}^2=\frac{\text{ia}^2}{\text{b}^2}=\text{I}$
Now according to Ampere’s circuital law, $\phi\text{B}\times\text{d}\ell=\text{B}\times2\times\pi\times\text{a}=\mu_0\text{I}$ $\Rightarrow\text{B}=\mu_0\frac{\text{ia}^2}{\text{b}^2}\times\frac{1}{2\pi\text{a}}=\frac{\mu_0\text{ia}}{2\pi\text{b}^2}$
Now according to Ampere’s circuital law, $\phi\text{B}\times\text{d}\ell=\text{B}\times2\times\pi\times\text{a}=\mu_0\text{I}$ $\Rightarrow\text{B}=\mu_0\frac{\text{ia}^2}{\text{b}^2}\times\frac{1}{2\pi\text{a}}=\frac{\mu_0\text{ia}}{2\pi\text{b}^2}$
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