Question

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

Answer 1

Answer: (C)

4 B

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, $\frac{1}{2}=2\pi r$

or $\frac{l}{4\pi }=r$

$\therefore B^{\prime }=\frac{{\mu }_0}{4\pi }.\frac{l.2\pi r\times 2}{r^2}$

Or $\therefore B^{\prime }=\frac{{\mu }_0}{4\pi }.\frac{l.\frac{1}{2}.2}{{\frac{l}{4\pi }}^2}$

Or $\therefore B^{\prime }=\frac{{\mu }_0}{4\pi }.\frac{ll\times 16{\pi }^2}{l^2}$ ....(ii)

Now, $B=\frac{{\mu }_0}{4\pi }.\frac{l.l}{{\frac{l}{2\pi }}^2}$ .....(iii)

Dividing Eq (ii) by Eq. (iii) we get

$\frac{\overline{B^{\prime }}}{B}=\frac{\frac{{\mu }_0}{4\pi }.\frac{I.l.16{\pi }^2}{l^2}}{\frac{{\mu }_0}{4\pi }.\frac{ll.4{\pi }^2}{l^2}}$

Or $\frac{\overline{B^{\prime }}}{B}=4$

Or $B^{\prime }=4B$