question_answer

A wire of length \[l\] is bent into a circular loop of radius R and carries a current\[I\]. The magnetic field at the centre of the loop is\[B\]. The same wire is now bent into a double loop of equal radii. If both loops carry the same current \[I\] and it is in the same direction, the magnetic field at the centre of the double loop will be

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 \[B=4B\]