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. \(2B\)
• C. \(4B\)
• D. \(8B\)

Hint:

If the same wire is divided into \(n\) loops, radius becomes \(R/n\) and total field becomes \(n^2B\).

Answer 1

The Correct Option is C

Step 1: Magnetic field at centre of a single loop.
\[ B = \frac{\mu_0 I}{2R} \]
Step 2: Wire bent into double loop.
Same wire length now forms two loops, so total length is divided into 2 equal circumferences.
Thus each loop has half the length, meaning radius becomes:
\[ 2\pi R' = \frac{1}{2}(2\pi R) \Rightarrow R' = \frac{R}{2} \]
Step 3: Field due to one smaller loop.
\[ B' = \frac{\mu_0 I}{2R'} = \frac{\mu_0 I}{2(R/2)} = \frac{\mu_0 I}{R} = 2B \]
Step 4: Total field due to two loops.
Both loops carry current in same direction, so fields add:
\[ B_{total} = 2B + 2B = 4B \]
Final Answer:
\[ \boxed{4B} \]