A wire of length $l$ is bent into a circular loop of radius $\mathrm{R}$ 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

  $2 \mathrm{~B}$
• Correct answer
  C

  $4 \mathrm{~B}$
• D

  $8 \mathrm{~B}$

Solution

At the centre of the loop, magnetic field $B=\frac{\mu_{0}}{4 \pi} \cdot \frac{I \cdot 2 \pi R}{R^{2}}$...(i)
For the wire which is looped double let radius becomes $r$ Then, $\frac{l}{2}=2 \pi r \quad$ or $\quad \frac{l}{4 \pi}=(r)$ $\therefore B^{\prime}=\frac{\mu_{0}}{4 \pi} \cdot \frac{I \cdot 2 \pi r \times 2}{r^{2}}$ or $B^{\prime}=\frac{\mu_{0}}{4 \pi} \cdot \frac{I \cdot \frac{l}{2} \cdot 2}{\left(\frac{1}{4 \pi}\right)^{2}}$ or $\quad B^{\prime}=\frac{\mu_{0}}{4 \pi} \cdot \frac{\Pi \times 16 \pi^{2}}{l^{2}}$...(ii) Now, $B=\frac{\mu_{0}}{4 \pi} \cdot \frac{I \cdot l}{\left(\frac{l}{2 \pi}\right)^{2}}\left[R=\frac{1}{2 \pi}\right]$...(iii) Dividing Eq. (ii) by(iii), $$ \frac{B^{\prime}}{B}=\frac{\frac{\mu_{0}}{4 \pi} \cdot \frac{I \cdot l \cdot 16 \pi^{2}}{l^{2}}}{\frac{\mu_{0}}{4 \pi} \cdot \frac{I l \cdot 4 \pi^{2}}{l^{2}}}=4 $$ or $B^{\prime}=4 B$

Hence, the correct answer is option: 'C'.