A flat coil has a maximum flux of $4.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$ through it, with its area perpendicular to a uniform magnetic field. When the coil is rotated $60^{\circ}$ (so that its normal makes a $60^{\circ}$ angle with the field), what is the flux through it?
A) $3.5 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$
B) $6.9 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$
C) $2.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$
D) $4.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$
E) $8.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$

Question

Quizplus · Accepted Answer

The magnetic flux through the coil is given by $\Phi = \Phi_{\text{max}} \cos \theta$. With $\Phi_{\text{max}} = 4.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$ and $\theta = 60^{\circ}$, the flux is $\Phi = 4.0 \times 10^{-3} \cos 60^{\circ} = 2.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$.

A Flat Coil Has a Maximum Flux Of 4.0×10−3 T⋅m24.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}4.0×10−3 T⋅m2

A Flat Coil Has a Maximum Flux Of $4.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}$