A) Zero
B) 2 B
C) 4 B
D) 8 B
Correct Answer: C
Solution :
Magnetic field at the centre of the loop \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{I.2\pi R}{{{R}^{2}}}\] ?..(i) For the wire which is looped double let radius becomes r Then, \[\frac{l}{2}=2\pi r\] or \[\frac{l}{4\pi }=(r)\] \[\therefore \] \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{I.\frac{l}{2}.2}{{{\left( \frac{l}{4\pi } \right)}^{2}}}\] or \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{ll\times 16{{\pi }^{2}}}{{{l}^{2}}}\] ?.(ii) Now, \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{I.l}{{{\left( \frac{l}{2\pi } \right)}^{2}}}\left[ R=\frac{l}{2\pi } \right]\] ?(iii) Dividing Eq. (ii) by Eq. (iii), we get \[\frac{B}{B}=\frac{\frac{{{\mu }_{0}}}{4\pi }.\frac{I.l.16{{\pi }^{2}}}{{{l}^{2}}}}{\frac{{{\mu }_{0}}}{4\pi }.\frac{Il.4{{\pi }^{2}}}{{{l}^{2}}}}\] or \[\frac{B}{B}=4\] or \[B=4B\]You need to login to perform this action.
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