Discussion
If the flux linked with the coil of area of cross-section 0.5 m2 placed in a magnetic field of 16 \(T\) is 4 Wb, then the angle between the magnetic field and the area vector of the coil is
Answer: D
The angle is 60°.
Magnetic flux through a flat coil is \(\phi = BA\cos\theta\), where \(\theta\) is the angle between \(\vec{B}\) and the area vector:
\[ \begin{aligned} 4 &= 16 \times 0.5 \times \cos\theta \\ \cos\theta &= \frac{4}{8} \\ &= \frac{1}{2}, \quad \theta \\ &= 60^{\circ} \end{aligned} \]
30° uses \(\sin\theta\), which is correct only if \(\theta\) is measured from the plane of the coil. 0° would give the maximum flux of 8 Wb, and 90° zero flux.
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