Magnetic Field Due to a Straight Wire
Category : JEE Main & Advanced
Magnetic field due to a current carrying wire at a point P which lies at a perpendicular distance r from the wire as shown is given as
\[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}\,(\sin {{\varphi }_{1}}+\sin {{\varphi }_{2}})\]
From figure \[\alpha =({{90}^{o}}-{{\varphi }_{1}})\]
and \[\beta =({{90}^{o}}+{{\varphi }_{2}})\]
Hence \[B=\frac{{{\mu }_{o}}}{4\pi }.\frac{i}{r}(\cos \alpha -\cos \beta )\]
(1) For a wire of finite length : Magnetic field at a point which lies on perpendicular bisector of finite length wire
\[{{\varphi }_{1}}={{\varphi }_{2}}=\varphi \]
So \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}(2\sin \varphi )\]
(2) For a wire of infinite length : When the linear conductor XY is of infinite length and the point P lies near the centre of the conductor \[{{\phi }_{1}}={{\phi }_{2}}={{90}^{o}}\].
So,\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{i}{r}[\sin {{90}^{o}}+\sin {{90}^{o}}]\]
\[=\frac{{{\mu }_{0}}}{4\pi }\frac{2i}{r}\]
(3) For a wire of semi-infinite length : When the linear conductor is of infinite length and the point P lies near the end Y or X. \[{{\varphi }_{1}}={{90}^{o}}\] and \[{{\varphi }_{2}}={{0}^{o}}\]
So, \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{i}{r}[\sin {{90}^{o}}+\sin {{0}^{o}}]\]
\[=\frac{{{\mu }_{0}}i}{4\pi r}\]
(4) For axial position of wire : When point P lies on axial position of current carrying conductor then magnetic field at P
\[B=0\]
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