If the strength of the current in a straight wire is doubled, how does the magnitude of the magnetic field at a fixed distance from the wire change?
The field strength doubles
The magnetic field around a long straight current-carrying wire is found from Ampère's circuital law, which relates the field integrated around a closed loop to the current enclosed. Applying it to a circular loop of radius r centred on the wire gives the standard result:
B = μ0I / (2πr)
Here μ0 is the permeability of free space, I is the current, and r is the perpendicular distance from the wire. At a fixed distance r, μ0 and 2πr are constants, so the field is directly proportional to the current: B ∝ I.
Doubling the current therefore doubles the field:
Bnew = μ0(2I) / (2πr) = 2 × [μ0I / (2πr)] = 2B
So the field strength doubles, which is the correct choice.
The other options misjudge the proportionality. The field cannot be halved or stay unchanged, since B rises in direct step with I, not inversely and not independently. It would only quadruple if B depended on I² (a square law) — but the relationship is strictly linear (first power of I), so a factor-of-two increase in current produces exactly a factor-of-two increase in field, not a factor of four.
Important Questions from Ampere's Circuit Law
Ampere circuital law states that :
Ampere's circuital law involves finding the _________.
