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…

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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

Aicharya · Accepted Answer

Magnetic field at the centre of the loop

For the wire which is looped double let radius becomes $r$ Then, $\begin{aligned} \frac{l}{2} & =2 \pi r \\ \text{or} \quad \frac{l}{4 \pi} & =(r) \end{aligned}$ $\begin{array}{ll}\therefore & B^{\prime}=\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot 2 \pi r \times 2}{r^2} \\ \text { or } & B^{\prime}=\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot \frac{l}{2} \cdot 2}{\left(\frac{l}{4 \pi}\right)^2}\end{array}$

Dividing Eq. (ii) by Eq. (iii), we get $\frac{B^{\prime}}{B}=\frac{\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot l \cdot 16 \pi^2}{l^2}}{\frac{\mu_0}{4 \pi} \cdot \frac{I l \cdot 4 \pi^2}{l^2}}$ $\begin{array}{ll}\text { or } & \frac{B^{\prime}}{B}=4 \\ \text { or } & B^{\prime}=4 B\end{array}$

A wire of length $l$ is bent into a circular loop of radius $R$ and carries a current $I$. The magnetic…

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