JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Magnetic Field Due to a Straight Wire

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

\[B=0\]