Q.A circular coil of 50 turns and area 0.20m2 is placed with its plane perpendicular to a uniform magnetic field of 0.4T, so that the normal to the coil is parallel to the field.
(a) Calculate the magnetic flux linked with the coil.
(b) If the coil is now turned so that its normal makes an angle of 60∘ with the field, find the new flux linked with it.
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✓ Free question
Concept understanding — Magnetic Flux
Definition. The magnetic flux ΦB through an area A in a field B is the number of field lines passing normally through that area, formally the surface integral ΦB=∫B⋅dA over the whole area, with the integral's dot product picking out only the field component along each element's own outward normal. For the common special case of a FLAT area A in a UNIFORM field B, making a fixed angle θ with the area's normal, this reduces to
ΦB=BAcosθ
Extremes. Flux is MAXIMUM (ΦB=BA) when the field is exactly along the area's normal (θ=0∘, cos0∘=1) -- i.e. when the field is perpendicular to the plane of the area itself. Flux is exactly ZERO when the field lies entirely IN the plane of the area (θ=90∘, cos90∘=0) -- i.e. when the area's normal is perpendicular to the field. A frequent source of numerical error is confusing these two angles: the angle a plane makes with the field, and the angle the plane's NORMAL makes with the field, always differ by exactly 90∘ from each other.
Unit. The SI unit is the tesla-metre-squared (T m2), given the named unit weber (Wb): 1Wb=1T m2. Flux is the single quantity whose CHANGE, via Faraday's law, is responsible for every induced emf covered in this unit -- whether that change comes from a varying field strength, a varying enclosed area, or a varying relative orientation (section 4.4).
"Magnetic flux formula and unit weber" and "magnetic flux class 12 important questions" are frequently searched terms tied to the Electromagnetic Induction chapter of the NCERT/CBSE Class 12 Physics curriculum, since flux is the foundational quantity behind every Faraday's-law numerical asked in board exams, JEE Main and NEET. The angle-confusion trap flagged here, between the field-to-plane angle and the field-to-normal angle, is one of the most common scoring errors in flux-based exam problems.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2026Set SEM31 markMCQ
Q.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 loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is
(a) 0·5 weber
(b) 1 weber
(c) 0 weber
(d) 2 weber
›Reveal solutionSolution
Magnetic flux φ = B·A·cos θ. The plane of the loop is perpendicular to B, so B is along the normal (θ = 0) and φ = BA = 0·5 × 1 = 0·5 Wb. Option (a).
Step 1 — area of the square loop: A = (1 m)² = 1 m².
Step 2 — orientation: 'plane of the loop perpendicular to B' means the field is along the area's normal, so the angle between B and the normal is 0° and cos θ = 1.
Step 3 — flux: φ = B A cos θ = 0·5 T × 1 m² × 1 = 0·5 Wb.
This flux definition underpins Faraday's law in the NCERT/CBSE Class 12 Physics chapter on Electromagnetic Induction.
✓Final answer
(a) 0·5 weber
CBSE 2023Set ANNUAL1 markMCQ
Q.The magnetic flux linked with a coil of N turns of area of cross-section A held with its plane parallel to the field B is
(1) NAB
(2) NAB/2
(3) NAB/4
(4) zero
›Reveal solutionSolution
Magnetic flux depends on the angle between the field and the AREA VECTOR (normal to the coil), not the plane itself.
Φ=NABcosθ, where θ is the angle between B and the normal n^ to the coil's plane. If the coil's plane is parallel to B, then the normal n^ is perpendicular to B, so θ=90∘ and cos90∘=0.
✓Final answer
(4) zero.
CBSE 2023Set ANNUAL1 mark
Q.Write the relation between weber and tesla units.
›Reveal solutionSolution
Tesla is the unit of magnetic flux density (B), weber is the unit of magnetic flux (Φ = B·A), so weber equals tesla times square metre.
Magnetic flux Φ through a surface of area A in a uniform field of magnetic flux density B is:
Φ=B⋅A
Here B is measured in tesla (T), the SI unit of magnetic flux density, and Φ is measured in weber (Wb), the SI unit of magnetic flux. Since Φ=BA, dimensionally:
1Wb=1T×1m2
Equivalently, 1T=1Wb/m2 — tesla is weber per square metre.
✓Final answer
1 weber = 1 tesla × 1 m² (Wb = T·m²), i.e. 1 T = 1 Wb/m².
CBSE 2016Set ANNUAL1 markMCQ
Q.Dimension of magnetic flux is
(a) [ML²T⁻²A⁻¹]
(b) [MLT⁻¹A⁻²]
(c) [ML⁻¹TA⁻¹]
(d) [MT⁻¹A]
›Reveal solutionSolution
Magnetic flux Φ = B·A; working out the dimension of B from F = qvB and multiplying by area gives [ML²T⁻²A⁻¹].
From F = qvB, the dimension of B = [F]/([q][v]) = (MLT⁻²)/((AT)(LT⁻¹)) = MT⁻²A⁻¹.