Q.A square coil of side 30 cm with 500 turns is kept in a uniform magnetic field of 0.4 T. The plane of the coil is inclined at an angle of 30∘ to the field. Calculate the magnetic flux through the coil.
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.
ΦB=NBAcosθ; with θ=90∘−30∘=60∘, ΦB=500(0.4)(0.09)(0.5)=9 Wb.
✓Final answer
ΦB=9 Wb.
Step 1. Given: N = 500 turns, side = 30 cm = 0.3 m so A=0.32=0.09m2, B=0.4 T, plane inclined 30∘ to the field.
Step 2. The angle used in ΦB=BAcosθ is between the field and the NORMAL, so θ=90∘−30∘=60∘.
Step 3. For N turns, total flux linkage is ΦB=NBAcosθ=500×0.4×0.09×cos60∘.