JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Biot-Savart's Law

Biot-Savart's law is used to determine the magnetic field at any point due to a current carrying conductor.

This law is although for infinitesimally small conductor yet it can be used for long conductors. In order to understand the Biot-Savart's law, we need to understand the term current-element.

Current element It is the product of current and length of infinitesimal segment of current carrying wire.

The current element is taken as  a vector quantity. Its direction is same as the direction of current. Current element \[AB=i\vec{d}l\]

According to Biot-Savart Law, magnetic field at point 'P' due to the current element \[i\vec{d}l\] is given by the expression, \[d\vec{B}=k\frac{i\,dl\sin \theta }{{{r}^{2}}}\hat{n}\] . Also \[\vec{B}=fd\vec{B}=\frac{{{\mu }_{0}}i}{4\pi }.\,\int_{{}}^{{}}{\frac{dl\sin \theta }{{{r}^{2}}}}\hat{n}\] 

In C.G.S. k = 1 and in S.I. \[k=\frac{{{\mu }_{0}}}{4\pi }\] 

where \[{{\mu }_{0}}=\] Absolute permeability of air or vacuum \[=4\pi \times {{10}^{-7}}\,\frac{Wb}{Amp-metre}\] . It's other units are 

\[\frac{henry}{metre}\]or \[\frac{N}{Am{{p}^{2}}}\] or \[\frac{Tesla-metre}{Ampere}\]

vectorially, \[d\vec{B}=\frac{{{\mu }_{0}}}{4\pi }.\frac{i(\vec{d}l\times \hat{r})}{{{r}^{2}}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{i(\vec{d}l\times \vec{r})}{{{r}^{3}}}\]

i.e., \[\vec{B}\] is perpendicular to both \[I\,\vec{d}\,l\] and

\[\vec{r}\]

.