Q. 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 -
BITSATBITSAT 2009
Solution:
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' = \frac{\mu_0}{4 \pi} . \frac{I.2 \pi r \times 2}{r^2} $
Or $B' = \frac{\mu_{0}}{4\pi} . \frac{I .\frac{l}{2}.2}{\frac{l}{4\pi}^{2}}$
Or $ B' = \frac{ \mu_{0}}{4\pi} . \frac{Il \times16 \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'}{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'}{B} = 4 $
Or $ B' = 4 B $
