JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Magnetic Field Due to an Infinite Sheet Carrying Current

The figure shows an infinite sheet of current with linear current density j (A/m). Due to symmetry the field line pattern above and below the sheet is uniform. Consider a square loop of side l as shown in the figure.

\[\int_{a}^{b}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{b}^{c}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{c}^{d}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{d}^{a}{\overrightarrow{B}.\overrightarrow{dl}}={{\mu }_{0}}i\]   (By Ampere's law)

Since \[B\bot dl\] along the path \[b\to c\] and \[d\to a,\] therefore, \[\int_{b}^{c}{\overrightarrow{B}.\overrightarrow{dl}}=0\]; \[\int_{d}^{a}{\overrightarrow{B}.\overrightarrow{dl}}=0\]

Also, \[B\,\,|\,\,|\,\,dl\] along the path \[a\to b\] and \[c\to d,\] thus \[\int_{a}^{b}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{d}^{a}{\overrightarrow{B}.\overrightarrow{dl}}=2Bl\]

The current enclosed by the loop is \[i=jl\]. Therefore, according to Ampere?s law \[2Bl={{\mu }_{0}}(jl)\] or \[B=\frac{{{\mu }_{0}}j}{2}\]