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# A long straight wire carries a current i. Let ${{B}_{1}}$ be the magnetic field at a point P at a distance d from the wire. Consider a section of length l of this wire such that the point P lies on a perpendicular bisector of the section. Let ${{B}_{2}}$ be the magnetic field at this point due to this section only. Find the value of $d/l$ so that ${{B}_{2}}$ differs from ${{B}_{1}}$ by 1%.

Last updated date: 05th Mar 2024
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Hint: Apply Biot Savart's law. Let us consider two magnetic fields ${{B}_{1}}\And {{B}_{2}}$.Given that the point P is at a distance d from the wire. Hence, here calculate the value of ${{B}_{1}}$. The value of ${{B}_{1}}$ is the product of permeability of free space and the current flowing through the wire to the product of two pi and the distance. Then calculate the magnetic field at ${{B}_{2}}$. Here also consider the same equation as the above. But here consider the sine of the angle also. Then by equating both ${{B}_{1}}\And {{B}_{2}}$we will get the value of $d/l$.

The magnetic field,
${{B}_{1}}=\dfrac{{{\mu }_{0}}I}{2\pi d}$
The magnetic field,
${{B}_{2}}=\dfrac{{{\mu }_{0}}I}{4\pi d}\left( \sin \theta +\sin \theta \right)$
$\Rightarrow {{B}_{2}}=\dfrac{{{\mu }_{0}}I}{4\pi d}\times 2\sin \theta$
$\Rightarrow {{B}_{2}}=\dfrac{{{\mu }_{0}}I}{4\pi d}\times 2\times \dfrac{l}{2\sqrt{{{d}^{2}}+{{\left( \dfrac{l}{2} \right)}^{2}}}}$
\begin{align} & \dfrac{{{B}_{1}}-{{B}_{2}}}{{{B}_{1}}}=0.01 \\ & \Rightarrow 1-\dfrac{{{B}_{2}}}{{{B}_{1}}}=0.01 \\ & \Rightarrow \dfrac{{{B}_{2}}}{{{B}_{1}}}=1-0.01 \\ & \Rightarrow \dfrac{{{B}_{2}}}{{{B}_{1}}}=0.99 \\ & \\ \end{align}
${{B}_{2}}={{B}_{1}}\times \dfrac{l}{2\sqrt{{{d}^{2}}+{{\left( \dfrac{l}{2} \right)}^{2}}}}$
$\Rightarrow 0.99=\dfrac{l}{2\sqrt{{{d}^{2}}+{{\left( \dfrac{l}{2} \right)}^{2}}}}$
$\Rightarrow {{0.99}^{2}}=\dfrac{{{l}^{2}}}{4\left( {{d}^{2}}+\dfrac{{{l}^{2}}}{4} \right)}$
$\Rightarrow {{0.99}^{2}}\times 4\times \left( {{d}^{2}}+\dfrac{{{l}^{2}}}{4} \right)={{l}^{2}}$
$\Rightarrow {{0.99}^{2}}\times \left( {{d}^{2}}+\dfrac{{{l}^{2}}}{4} \right)=\dfrac{{{l}^{2}}}{4}$
$\Rightarrow {{0.99}^{2}}{{d}^{2}}=\dfrac{{{l}^{2}}}{4}\left( 1-{{0.99}^{2}} \right)$
$\Rightarrow {{0.99}^{2}}{{d}^{2}}=\dfrac{{{l}^{2}}}{4}\times 0.0199$
$\Rightarrow 197.005{{d}^{2}}={{l}^{2}}$
Rearranging the equation and taking the square root we get,
\begin{align} & \Rightarrow \dfrac{{{d}^{2}}}{{{l}^{2}}}=\dfrac{1}{197.005} \\ & \therefore \dfrac{d}{l}=0.0712 \\ \end{align}