A) \[\frac{B}{4}\]
B) \[\frac{B}{2}\]
C) 2 B
D) 4 B
Correct Answer: D
Solution :
Magnetic field at the centre of circular coil \[B=\frac{{{\mu }_{0}}Ni}{2r}\] Ist case: \[N=1,\,L=2\pi r\Rightarrow r=\frac{L}{2\pi }\] \[\therefore \] \[B=\frac{{{\mu }_{0}}\times 1\times i}{2r}=\frac{{{\mu }_{0}}i}{2r}\] IInd Case: \[N=2,\,L=2\times 2\pi r'\] \[\Rightarrow \] \[r'=\frac{L}{4\pi }=\frac{r}{2}\] \[\therefore \] \[B'=\frac{{{\mu }_{0}}\times 2\times i}{2r'}\] \[=\frac{{{\mu }_{0}}\times 2i}{2\times (r/2)}=\frac{4{{\mu }_{0}}i}{2r}=4B\] Note: Magnetic field at the centre of circular coil is maximum and decreases as we move away from the centre (on the axis of coil)
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