Q.Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil?
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Magnetic Force Balance
When a current-carrying wire or coil sits in a magnetic field, it feels a force F=BILsinθ (or, for a point charge, F=qvBsinθ). On its own that force just pushes the conductor - but in many real situations the push is deliberately set up to CANCEL another force, so the whole system sits in equilibrium. That equilibrium condition - magnetic force balanced against weight, against another wire's magnetic force, or against a mechanical counterweight - is what "magnetic force balance" means, and it is also historically how the ampere itself was defined.
The balance condition
Whenever a conductor is in equilibrium under a magnetic force and one other force, the two must be equal and opposite:
BILsinθ=Fother
Solving this equation for whichever quantity is unknown (B, I, L, or the other force) is the entire skill in this class of problem - the only new step, beyond the force law itself, is correctly identifying what the magnetic force is opposing.
Case 1: a wire suspended against gravity
A straight horizontal wire of mass m and length l, carrying current I, can be held up ("floated") in mid-air by a horizontal magnetic field perpendicular to it. The upward magnetic force must equal the downward weight:
BIl=mg⟹B=Ilmg
For example, a 200g, 1.5m wire carrying 2A needs B=(2)(1.5)(0.2)(9.8)≈0.65T to stay suspended.
Case 2: two wires balancing each other
Two long parallel wires carrying currents I1,I2 exert a force per unit length on each other of 2πdμ0I1I2 (attractive if the currents run the same way, repulsive if opposite). If one wire is free to move, this magnetic force can itself balance that wire's weight:
2πhμ0I2L=mg⟹h=2πmgμ0I2L
This is exactly how a "current balance" apparatus works, and historically it is how the ampere was defined: the current that, flowing in two infinite parallel wires one metre apart, produces a force of exactly 2×10−7N per metre of length.
Case 3: balancing on a beam
A current-carrying coil arm hanging from one pan of a beam balance feels an extra force F=NBIl when only that arm sits in an external field. Re-balancing the beam means adding a mass m so that mg=NBIl.
Always check which length enters the formula - for a coil of N turns the force multiplies by N; for a single suspended straight wire it doesn't. …
Why this formula?
Magnetic Force Balance: Why the Key Formulas Hold
The Magnetic Force Balance describes when the magnetic force on a charged particle or current-carrying conductor is exactly balanced by another force (gravity, electric force, or tension). Let's build the reasoning step-by-step.
1. The Core Idea: What Does "Balance" Mean?
A force balance means the net force on an object is zero:
Fnet=0
For magnetic forces we use the Lorentz force law:
- On a moving charge: Fm=q(v×B)
- On a current-carrying wire: Fm=I(L×B)
When this is balanced by another force (say gravity Fg=mg):
Fm+Fother=0
2. Case 1: Charged Particle in Crossed Fields (Velocity Selector)
A charged particle moves perpendicular to both electric field E and magnetic field B.
- Electric force: Fe=qE (along E)
- Magnetic force: Fm=q(v×B) (perpendicular to both v and B)
For straight-line motion (no deflection), the two forces must cancel:
qE=qvB⇒v=BE
Key insight: Only particles with this exact speed pass undeflected — this is how velocity selectors work in mass spectrometers.
3. Case 2: Current-Carrying Wire Balanced by Gravity
A horizontal wire carrying current I sits in a perpendicular magnetic field B, suspended by strings.
The magnetic force on a straight wire is Fm=ILBsinθ; for a wire perpendicular to the field (θ=90∘), Fm=ILB. Setting this equal to the weight Fg=mg for equilibrium:
ILB=mg
Key insight: This balance lets you measure B if I, L, and m are known — the principle behind a current balance experiment.
4. Case 3: Circular Motion of a Charged Particle …
The key idea is the magnetic field at the centre of a circular current-carrying coil, given by the formula derived from the Biot–Savart law.
For a single circular loop of radius R carrying current I, the field at the centre is B=2Rμ0I. For a coil with N tightly wound turns, the fields of each turn add directly, so:
B=N⋅2Rμ0I
Here, N=100, I=1 A, R=10 cm=0.1 m, and μ0=4π×10−7 T m/A.
Substitute: …
The magnetic field at the centre of a circular coil is given by B=2Rμ0NI. For N=100, I=1 A, R=0.1 m, the magnitude is 6.28×10−4 T.
Why the Biot–Savart Law?
The magnetic field at the centre of a current-carrying circular loop arises from the Biot–Savart law. Each tiny current element Idl on the loop produces a magnetic field at the centre that points along the axis (perpendicular to the plane of the loop). Because of symmetry, the contributions from all elements add constructively -- there is no cancellation. The key insight: every element is at the same distance R from the centre, and the angle between dl and the radial vector is always 90∘, so the cross product simplifies beautifully.
B=2Rμ0NI
This is the central result for a tightly wound coil of N turns. Let's derive it step by step.
Step-by-step solution
- Start with a single turn. For a single circular loop of radius R, carrying current I, the magnetic field at the centre is
B1=2Rμ0I
This comes from integrating the Biot–Savart law: each element Idl contributes dB=4πμ0r2Idlsinθ. Here r=R and θ=90∘ (since dl is tangent and the vector from element to centre is radial), so sinθ=1. The integral ∮dl=2πR gives the result.
- Account for multiple turns. The coil has N=100 turns, all tightly wound so they essentially occupy the same radius R. The fields from each turn add linearly (superposition). Hence
B=N⋅B1=2Rμ0NI
-
Plug in the numbers.
- μ0=4π×10−7 T⋅m/A
- N=100
- I=1 A
- R=10 cm=0.1 m
So …
Method: Magnetic Field at the Centre of a Circular Coil (Biot–Savart Law)
This is a standard application of the Biot–Savart law for a circular current loop. The symmetry of the problem allows a direct formula.
Steps
- Recall the formula For a single circular loop of radius R carrying current I, the magnetic field at the centre is:
Bsingle=2Rμ0I
where μ0=4π×10−7 T m/A.
- Account for multiple turns For a tightly wound coil of N turns, each turn contributes the same field at the centre (since they are concentric). The total field is simply:
Btotal=N⋅2Rμ0I
- Substitute the given values
- N=100
- I=1 A
- R=10 cm=0.1 m
B=100×2×0.1(4π×10−7)×1
- Simplify step-by-step
- Denominator: 2×0.1=0.2
- Numerator: 4π×10−7
- So: …
Here are the most common mistakes students make when solving this problem, along with how to avoid each one.
1. Forgetting the Number of Turns (N)
The Mistake:
Students often use the formula for a single loop (B=2Rμ0I) and forget to multiply by the number of turns N.
Why it happens:
The problem says "tightly wound 100 turn coil." A coil is not a single loop — each turn contributes to the field at the centre.
How to Avoid:
Always check: Is it a single loop or a coil with N turns?
For a coil, the correct formula is:
B=2Rμ0NI
Here, N=100, so the field is 100 times that of a single loop.
2. Using the Wrong Radius Units
The Mistake:
Plugging R=10 directly into the formula without converting to metres.
Why it happens:
The radius is given in cm, but the SI unit for R in the formula is metres.
How to Avoid:
Always convert cm to m before substituting:
R=10 cm=0.1 m
If you forget, your answer will be off by a factor of 10.
3. Confusing the Formula for a Solenoid vs. a Circular Coil
The Mistake:
Using B=μ0nI (the solenoid formula) instead of the circular coil formula.
Why it happens:
Both involve "coils" and "turns," but the geometry is different. A solenoid is long and straight; a circular coil is a flat loop.
How to Avoid:
- Circular coil (flat): B=2Rμ0NI (centre field)
- Solenoid (long): B=μ0nI (inside field, n = turns per unit length)
If the problem says "radius" and "centre," it's a circular coil.
4. Forgetting the Direction of the Magnetic Field
The Mistake:
Only calculating magnitude but ignoring that the question asks for "magnitude" — some students still try to assign a direction incorrectly.
Why it happens:
Direction is important in many problems, but here only magnitude is asked. Overthinking can lead to errors.
How to Avoid: …
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.The radius of a coil of wire with N turns is 0.1 m and 2A current flows in the coil as shown. A long straight wire carrying a current of 20π A as shown is located at 0.5 m from the centre of the coil. The number of turns in the coil if the resultant magnetic field at the centre of the coil is zero [FIGURE: a circular coil of radius r carrying a current of 2A, with a current-direction arrow shown on the loop; a long straight wire is drawn above the coil (dashed line) carrying a current of 20π A shown flowing to the left, positioned at a perpendicular distance of 0.5 m from the coil's centre] (A) 2 (B) 4 (C) 6 (D) 10
›Reveal solutionSolution
This tests superposition of the magnetic field of a circular coil (at its centre) and of a long straight wire, set to cancel — solve for N by equating magnitudes.
Concept and Intuition
A circular coil of N turns carrying current I produces a field at its own centre of B=2rμ0NI, directed along the coil's axis (direction fixed by the right-hand rule for the shown current sense). A long straight wire carrying current I produces, at perpendicular distance d, a field B=2πdμ0I, circling the wire (again right-hand rule). The problem is engineered so that, given the current directions in the figure, these two fields point in opposite directions at the coil's centre. "Resultant field is zero" therefore just means the two magnitudes are equal — the geometry/direction part is already built into the problem statement, so we only need magnitude balance.
Step-by-Step Solution
- Field due to the coil at its centre: Bcoil=2rμ0NIcoil=2(0.1m)μ0N(2)=10μ0N. …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.If a straight current carrying wire of linear density 0.12 kgm−1 is suspended in mid air by a uniform horizontal magnetic field of 0.5 T normal to the length of the wire, then the current through the wire is (Acceleration due to gravity =10 ms−2; Neglect earth's magnetic field) (A) 2.4 A (B) 1.2 A (C) 0.6 A (D) 4.8 A
›Reveal solutionSolution
A current-carrying wire floats in a horizontal magnetic field when the magnetic force exactly cancels gravity; solving gives I=2.4 A.
Concept and Intuition
A straight wire carrying current I in a magnetic field B (perpendicular to the wire) experiences a force per unit length F/L=BI. For the wire to be suspended in mid-air (in equilibrium), this magnetic force must balance the weight per unit length of the wire, which is λg where λ is the linear mass density.
Step-by-Step Solution
- Force balance per unit length: BI=λg.
- Solve for current: I=Bλg.
- Substitute values: λ=0.12 kg m−1, g=10 m s−2, B=0.5 T. …
- AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQQ.Two infinitely long wires are placed at (1cm, 1cm) and (+1cm, -1cm) with 1A current in each and in the same directions perpendicular to x-y plane. Let the magnetic field due to these current carrying wires at the origin be B. If B0 is the magnitude of the field if only one of them was present, then B0∣B∣ is (A) 2 (B) 1 (C) 21 (D) 221
›Reveal solutionSolution
Two parallel wires symmetric about the x-axis add their fields at the origin constructively along one direction; the resultant is 2 times the field of either wire alone.
Concept and Intuition
An infinite straight wire carrying current I produces a field of magnitude μ0I/(2πd) at perpendicular distance d, circling the wire (direction given by z^×r^, where r^ points from the wire towards the field point, for current along +z^). With two wires we must add the two field vectors, not just their magnitudes.
Step-by-Step Solution
- Wire 1 is at (1,1) cm, wire 2 at (1,−1) cm; both distances from the origin are d=12+12=2 cm, so each alone gives a field of magnitude B0=2π2μ0I.
- Vector from wire 1 to origin: r1=(−1,−1). Field direction ∝z^×r1=(1,−1,0) (up to normalization).
- Vector from wire 2 to origin: r2=(−1,1). Field direction ∝z^×r2=(−1,−1,0). …
- AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQQ.Two long parallel straight metal wires A and B carrying currents 12 A and 36 A respectively, in the same direction are separated by 50 cm. The point relative to A, where the resultant magnetic induction between the two wires due to the currents is zero, will be (A) 90 cm (B) 7.5 cm (C) 28 cm (D) 12.5 cm
›Reveal solutionSolution
With both currents in the same direction, the magnetic fields cancel only in the region between the two wires. Equating BA=BB and solving gives the null point at 12.5 cm from wire A (closer to the weaker current, as expected).
Concept and Intuition
Each long straight wire produces a field B=2πdμ0I circling around it. Between two wires carrying current in the same direction, the two fields point in opposite directions (one wire's field goes into the page there, the other's comes out), so they can cancel at some point between them. The cancellation point sits closer to the wire with the smaller current (since a weaker source needs to be closer to match the stronger one's field at the same magnitude).
Step-by-Step Solution
- Let the null point be at distance x from wire A, so it is at (50−x) cm from wire B.
- Equate the magnitudes: 2πxμ0(12)=2π(50−x)μ0(36).
- Cancel common factors: x12=50−x36. …
- AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQQ.A tangent galvanometer has a coil of 50 turns and a radius of 20 cm. The horizontal component of earth's magnetic field is 3×10−5 T. What will be the current which gives a deflection of 45°? (A) 5π3 A (B) 3π5 A (C) 53π A (D) 35π A
›Reveal solutionSolution
At a 45° deflection the tangent law gives tan45°=1, so the coil's magnetic field exactly equals Earth's horizontal field, letting us solve directly for the current: I=5π3A.
Concept and Intuition
A tangent galvanometer balances the magnetic field of its coil against Earth's horizontal field H; the needle's deflection θ obeys tanθ=HBcoil, where Bcoil=2rμ0nI.
Step-by-Step Solution
- Tangent law: Bcoil=Htanθ. At θ=45°, tan45°=1, so Bcoil=H.
- Bcoil=2rμ0nI, so 2rμ0nI=H⇒I=μ0n2rH.
- Substitute r=0.2m, H=3×10−5T, n=50, μ0=4π×10−7: …
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