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

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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

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At the centre of the loop, magnetic field $B = \frac{μ _{0}}{4 π} \cdot \frac{I \cdot 2 π R}{R ^{2}}$ ...(i)
For the wire which is looped double let radius becomes $r$ Then, $\frac{l}{2} = 2 π r$ or $\frac{l}{4 π} = (r)$ $∴ B^{'} = \frac{μ _{0}}{4 π} \cdot \frac{I \cdot 2 π r \times 2}{r ^{2}}$ or $B^{'} = \frac{μ _{0}}{4 π} \cdot \frac{I \cdot \frac{l}{2} \cdot 2}{( \frac{1}{4 π} ) ^{2}}$ or $B^{'} = \frac{μ _{0}}{4 π} \cdot \frac{Π \times 16 π ^{2}}{l ^{2}}$ ...(ii) Now, $B = \frac{μ _{0}}{4 π} \cdot \frac{I \cdot l}{( \frac{l}{2 π} ) ^{2}} [R = \frac{1}{2 π}]$ ...(iii) Dividing Eq. (ii) by(iii), $\frac{B ^{'}}{B} = \frac{\frac{μ _{0}}{4 π} \cdot \frac{I \cdot l \cdot 16 π ^{2}}{l ^{2}}}{\frac{μ _{0}}{4 π} \cdot \frac{I l \cdot 4 π ^{2}}{l ^{2}}} = 4$ or $B^{'} = 4 B$

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