Magnetic Flux Formula Φ = BA cosθ: How to Use It

Physics · Electromagnetic Induction · NEET

Magnetic flux is found from Φ = BA cosθ, where B is the field, A is the loop area, and θ is the angle between B and the loop's NORMAL (the line pointing straight out of the surface), not the loop's plane. When B is perpendicular to the plane (points along the normal), θ = 0, cosθ = 1, and flux is maximum: Φ = BA. Memory hook: "theta is measured from the arrow poking OUT of the loop, not from the surface."
Magnetic Flux: Φ = B A cosθLoop, area AnormalBθθ = 0° (plane ⊥ B):Φ = BA (MAX)θ = 90° (plane ∥ B):Φ = 0θ = angle B–normal
θ in Φ = BA cosθ is the angle between B (red) and the loop's normal (black arrow), not the loop's plane. When the plane is perpendicular to B the normal lines up with B (θ = 0°, flux maximum); when the plane is parallel to B the normal is perpendicular to B (θ = 90°, flux zero).

Your doubts, answered

Is θ measured from the plane of the loop or from the normal?

Always from the NORMAL — the imaginary arrow that points straight out of the flat loop. This is the single most common mistake. If a question gives you the angle with the PLANE, convert it: angle with normal = 90° minus angle with plane. So if B makes 30° with the plane, it makes 60° with the normal, and you use cos 60° = 0.5.

When the plane of the loop is perpendicular to B, what is θ?

θ = 0°, so cosθ = 1 and flux is MAXIMUM (Φ = BA). If the plane is perpendicular to B, the normal is parallel to B, so the angle between them is zero. NEET 2022 tested exactly this wording: 'plane perpendicular to B' gave Φ = BA, not zero.

When is magnetic flux zero?

Flux is zero when the loop's PLANE is PARALLEL to B (field lines slide along the surface and none pass through). Then the normal is perpendicular to B, θ = 90°, cos 90° = 0, so Φ = 0. Do not confuse 'plane parallel to B' (flux = 0) with 'plane perpendicular to B' (flux = max).

What exactly is A in Φ = BA cosθ?

A is the flat area enclosed by the loop, in square metres. For a square of side L, A = L². For a circle of radius r, A = πr². If there are N turns, each turn has the same area A — flux per turn is still BA cosθ; you multiply by N only later in the EMF or flux-linkage formula (NΦ).

What if B is not uniform over the loop?

Then Φ = BA cosθ only works for the part where B is present. A classic trap: a loop bigger than the field region encloses flux only over the FIELD's area, not the loop's area. Use the area where the field actually exists (πr² of the field patch), not the loop's larger area.

⚠️ The NEET trap
The question says 'plane of the loop is perpendicular to B', so the angle is 90° and cosθ = 0, giving zero flux.
'Plane perpendicular to B' means the NORMAL is parallel to B, so θ = 0° and cosθ = 1, giving maximum flux Φ = BA. θ is always the angle with the normal, never with the plane.
🧠 Plane ⊥ B → flux MAX. Plane ∥ B → flux ZERO. Read whether they name the plane or the normal, then flip if needed.

Real NEET questions

NEET 2022

A square loop of side 1 m and resistance 1 Ω is placed in a magnetic field of 0.5 T. If the plane of the loop is perpendicular to the direction of the magnetic field, the magnetic flux through the loop is:

A · 2 weber
B · 0.5 weber
C · 1 weber
D · Zero weber
Solution: Step 1 — Formula: Φ = BA cosθ, where θ is the angle between B and the loop's NORMAL. Step 2 — Read the geometry: 'plane perpendicular to B' means B lies along the normal, so θ = 0° and cosθ = 1. Step 3 — Area: side 1 m, so A = 1 × 1 = 1 m². Step 4 — Substitute: Φ = 0.5 × 1 × 1 = 0.5 Wb. Answer: B. (Resistance 1 Ω is extra information — flux does not depend on resistance.)
NEET 2019

A 800 turn coil of effective area 0.05 m² is kept perpendicular to a magnetic field 5 × 10⁻⁵ T. When the plane of the coil is rotated by 90° about one of its coplanar axes in 0.1 s, the emf induced in the coil will be:

A · 2 V
B · 0.2 V
C · 2 × 10⁻³ V
D · 0.02 V
Solution: Step 1 — Initial flux: 'coil kept perpendicular to B' → plane ⊥ B → θ = 0° → Φ_i = NBA cos0 = NBA (maximum). Step 2 — Final flux: after a 90° rotation the plane becomes parallel to B → θ = 90° → cos90° = 0 → Φ_f = 0. Step 3 — Average emf = |ΔΦ|/Δt = NBA/Δt. Step 4 — Substitute: (800 × 5×10⁻⁵ × 0.05) / 0.1 = (2×10⁻³)/0.1 = 0.02 V. Answer: D. This shows how cosθ turns a full flux into zero when the angle swings from 0° to 90°.

Solved Electromagnetic Induction NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 16 Electromagnetic Induction NEET PYQs ›
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Frequently asked

What is the SI unit of magnetic flux?

The weber (Wb). From Φ = BA cosθ, unit = tesla × metre² = T·m² = Wb. One weber is one tesla times one square metre.

Does θ depend on how I orient the normal?

You may point the normal either way out of the loop; cosθ then differs only in sign. For magnitude of flux (as NEET asks), take the acute angle so cosθ is positive. The sign matters only when you later track the DIRECTION of induced EMF using Lenz's law.

Is Φ = BA cosθ the same as Φ = B·A (dot product)?

Yes. B·A is the vector dot product of the field and the area vector (area vector points along the normal, magnitude A). The dot product equals BA cosθ, so both forms are identical.

What is the difference between flux and flux linkage?

Flux Φ = BA cosθ is for a single turn. Flux linkage is NΦ for N turns. You use flux linkage (NΦ) in EMF and inductance formulas, but the basic Φ = BA cosθ describes just one loop.

Why is this formula important for NEET?

Almost every EMI numerical starts by computing flux with Φ = BA cosθ. Faraday's law (EMF = −N dΦ/dt), AC generators, and inductance all build on it. Getting θ right (normal vs plane) is the single biggest source of lost marks in this chapter.