Concept:Magnetic flux is the dot product of the magnetic field vector and the area vector. Care must be taken depending on whether the angle is given relative to the surface plane or the normal vector.
Historical Note: In many past papers from this region, "angle with the field" is interpreted as the angle with the area vector unless specified as the plane.Formula:$$ \Phi = BA \cos(\theta) $$
Solution:- Identify parameters: \( B = 0.5 \ \text{T} \), \( A = 2 \ \text{m}^2 \).
- Use the angle provided exactly as \( \theta = 60^\circ \) .
- Recall that \( \cos(60^\circ) = 0.5 \).
- Calculate: \( \Phi = (0.5)(2) \cos(60^\circ) = 1 \times 0.5 = 0.5 \ \text{Wb} \).
Why other options are incorrect:Option A has the incorrect unit (Tesla is for field, not flux). Options C and D calculate flux using a sine or a \( 30^\circ \) angle, or just have incorrect math/units.
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Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.