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 $B$. 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)

Magnetic field at the centre of the loop
$B=\frac{u_0}{4\pi }.\frac{I.2\pi R}{R^2}$ ......(i)
For the wire which is looped double let radius becomes r
Then , $\frac{1}{2}=2\pi r$
or $\frac{l}{4\pi }=r$
$\therefore B^{\prime }=\frac{{\mu }_0}{4\pi }.\frac{I.2\pi r\times 2}{r^2}$
Or $B^{\prime }=\frac{{\mu }_0}{4\pi }.\frac{I.\frac{l}{2}.2}{{\frac{l}{4\pi }}^2}$
Or $B^{\prime }=\frac{{\mu }_0}{4\pi }.\frac{Il\times 16{\pi }^2}{l^2}$ ...(i)
Now, $B=\frac{{\mu }_0}{4\pi }.\frac{I.l}{{\frac{l}{2\pi }}^2}R=\frac{l}{2\pi }$ ...(ii)
Dividing Eq (ii) by Eq. (iii) we get
$\frac{B^{\prime }}{B}=\frac{\frac{{\mu }_0}{4\pi }.\frac{I.l.16{\pi }^2}{l^2}}{\frac{{\mu }_0}{4\pi }.\frac{Il.4{\pi }^2}{l_2}}$
Or $\frac{B^{\prime }}{B}=4$
Or $B^{\prime }=4B$