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A long straight wire of radius $R$ carries current $i$. The magnetic field inside the wire at distance $r$ from its centre is expressed as :
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Verified Answer
The correct answer is:
$\left(\frac{\mu_{0} i}{2 \pi R^{2}}\right) \cdot r$
Using Ampere's law, we have
$\begin{array}{l}
\oint \vec{B} \cdot d \vec{\ell}=\mu_{0} i_{i n} \\
\text { or } \quad B \times 2 \pi r=\mu_{0} \frac{i}{\pi R^{2}} \pi r^{2} \\
\therefore \quad \mathrm{B}=\frac{\mu_{0}}{2 \pi} \cdot \frac{i r}{R^{2}}
\end{array}$
$\begin{array}{l}
\oint \vec{B} \cdot d \vec{\ell}=\mu_{0} i_{i n} \\
\text { or } \quad B \times 2 \pi r=\mu_{0} \frac{i}{\pi R^{2}} \pi r^{2} \\
\therefore \quad \mathrm{B}=\frac{\mu_{0}}{2 \pi} \cdot \frac{i r}{R^{2}}
\end{array}$
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