Physics · Moving Charges And Magnetism · NEET
There are two different situations. If you have N separate coils all of the SAME radius r, then B = μ₀NI/(2r), which is simply N times one loop. But in the bending problem you take ONE wire of fixed length and rebend it. Making n turns forces each loop to shrink to radius r/n. So you get one factor n from the turns AND another factor n from the smaller radius, giving n². Always ask: is the radius fixed (then factor n) or is the wire length fixed (then factor n²)?
The total wire length does not change when you rebend it. For 1 loop the length is 2πr. For n identical loops the length is n × 2πr'. Setting them equal: 2πr = n·2πr', so r' = r/n. The wire is shared among n circles, so each circle is n times smaller.
No. The problem says a steady current I flows through the same wire from the same source, so I stays the same. Only the geometry (number of turns and radius) changes. Do not scale the current.
Start with B = μ₀NI/(2r). For n turns put N = n and r → r/n: B_new = μ₀·n·I / (2·(r/n)) = μ₀·n·I·n / (2r) = n²·(μ₀I/2r) = n²B. The n on top comes from turns, and the n from cancelling the r/n in the denominator.
Put n = 2. New field = 2²B = 4B. For 3 turns it is 9B, for 4 turns 16B. The field grows very fast because of the squared factor, which is why tightly wound multi-turn coils give strong fields.
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be:
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take μ₀ = 4π × 10⁻⁷ SI units):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B = μ₀NI/(2R), where N is the number of turns, I is the current, R is the radius of each turn, and μ₀ = 4π × 10⁻⁷ T·m/A. This works when all N turns have the same radius R and are stacked at the same centre.
Because rebending keeps the wire length fixed. n turns means each loop shrinks to radius r/n. One factor of n comes from more turns and a second factor of n comes from the smaller radius, giving n² overall.
Yes, but it changes differently. Magnetic moment m = NIA = n·I·π(r/n)² = πIr²/n, so the moment actually DECREASES as 1/n while the central field INCREASES as n². Do not mix up the two results.
No. A solenoid stacks turns along a length so its field is μ₀nI (n = turns per unit length) and is nearly uniform inside. The N-turn flat coil here has all turns at one point, giving B = μ₀NI/(2R) only at the centre.
