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Question:

Two identical wires A and B, each of length 'l', carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side 'a'. If BA and BB are the values of magnetic field at the centres of the circle and square respectively, then the ratio BA/BB is.

π/16

π/16√2

π/8√2

π/8

Solution:

For wire A:
2πR = l ⇒ R = l/2π
Magnetic Field at the centre of ring is given by:
BA = μ0I/2R
For wire B:
4a = l; a = l/4;
Magnetic field due to wire of of length L at a distance d is given by:
BB = μ0I/4πd(cosθ1 + cosθ2) ⇒ BB = 2√2μ0I/πl
BBnet at the centre of the square is: 8√2μ0I/πl (Due to all four wires)
Now, BA/BB = π/8√2