When a current-carrying conductor is bent into a loop, it produces a magnetic field. The strength of this magnetic field depends on the number of turns in the loop, the current flowing through the wire, and the distance from the center of the loop.


When the conductor is bent into a loop with one turn, the magnetic field produced at the center of the loop can be calculated using the formula:

B = (μ₀ * I) / (2 * r)

Where B is the magnetic field, μ₀ is the permeability of free space, I is the current flowing through the wire, and r is the radius of the loop.


When the conductor is bent into a loop with two turns, the magnetic field produced at the center of the loop can be calculated using the formula:

B = (μ₀ * I * N) / (2 * r)

Where B is the magnetic field, μ₀ is the permeability of free space, I is the current flowing through the wire, N is the number of turns in the loop, and r is the radius of the loop.


Comparing the two formulas, we can see that the magnetic field produced by the loop with two turns is directly proportional to the number of turns in the loop. Therefore, the magnetic field at the center of the loop with two turns will be stronger than the magnetic field at the center of the loop with one turn.


In conclusion, the magnetic field produced by a current-carrying conductor that is bent into a loop with two turns is stronger than the magnetic field produced by a loop with one turn. This is due to the fact that the magnetic field is directly proportional to the number of turns in the loop.

L= 2Rπ, ,R=l/2πMagnetic field =1×µi×2π/2l =µiπ/lIn second case, l=2×2πr, r=l/4πSo, magnetic field=2×i4πµ/l×2 =4µπi/lSo, B/B'=1/4