Magnetic Field at Centre O in Different Conditions of Circular Current
Category : JEE Main & Advanced
Condition | Figure | Magnetic field |
Arc subtends angle \[\theta \] at the centre | \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{\theta \,i}{r}\] | |
Arc subtends angle \[(2\pi -\theta )\]at the centre | \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{(2\pi -\theta )\,i}{r}\] | |
Semi-circular arc | \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{\pi i}{r}=\frac{{{\mu }_{0}}i}{4r}\] | |
Three quarter semi-circular current carrying arc | \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{\left( 2\pi -\frac{\pi }{2} \right)\,i}{r}\] \[=\frac{3{{\mu }_{0}}i}{8r}\] | |
Circular current carrying arc | \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi i}{r}\] \[=\frac{{{\mu }_{0}}i}{2r}\] | |
Concentric co-planer circular loops carries current in the same direction | \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }2\pi i\,\left( \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}} \right)\] | |
Concentric co-planer circular loops carries current in the opposite direction | \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }2\pi i\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\] | |
Concentric loops but their planes are perpendicular to each other | \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}\] \[=\frac{{{\mu }_{0}}}{2r}\sqrt{i_{1}^{2}+i_{2}^{2}}\] | |
Concentric loops but their planes are at an angle q with each other | \[B=\sqrt{\begin{align} & B_{1}^{2}+B_{2}^{2} \\ & +2{{B}_{1}}{{B}_{2}}\cos \theta \\ \end{align}}\] | |
Distribution of current across the diameter | \[B=0\] | |
Distribution of current between any two points on the circumference | \[B=0\] |
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