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)

4 B

Answer 2

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 E (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 $B' = 4 B$