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)You need to login to perform this action.
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