Question 18.QE.2: Flux through a book cover A book is positioned in a uniform ......
Flux through a book cover
A book is positioned in a uniform 0.20-T \vec{B} field that points from left to right in the plane of the page, shown in the figures on the next page. (For simplicity, we depict the book as a rectangular loop.) Each side of the book’s cover measures 0.10 m. Determine the magnetic flux through the cover when (a) the cover is in the plane of the page (figure a), (b) the cover is perpendicular to the plane of the page and the normal vector makes a 60° angle with the \vec{B} field (figure b), and (c) the book’s cover area is perpendicular to the plane of the page and the normal vector points toward the top of the page (figure c).
Represent mathematically The magnetic flux through an area is determined by Eq. (18.1) Φ = BA cos θ. The angle in each of the three situations is (a) θ = 90°, (b) θ = 60°, and (c) θ = 90°.
Solve and evaluate The magnetic flux through the book cover in each case is
\text { (a) } \Phi=(0.20 \mathrm{~T})(0.10 \mathrm{~m})^2 \cos \left(90^{\circ}\right)=0
\text { (b) } \Phi=(0.20 \mathrm{~T})(0.10 \mathrm{~m})^2 \cos \left(60^{\circ}\right)=1.0 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^2
\text { (c) } \Phi=(0.20 \mathrm{~T})(0.10 \mathrm{~m})^2 \cos \left(90^{\circ}\right)=0
We can evaluate these results by comparing the calculated fluxes to the number of \vec{B} field lines through the book cover’s area. Note that for the orientation of the book in (a) and (c), the \vec{B} field lines are parallel to the book’s area and therefore do not go through it. Those positions are consistent with our mathematical result. The orientation for the book in (b) is such that some \vec{B} field lines do pass through the book, which is also consistent with the nonzero mathematical result.
Try it yourself: A circular ring of radius 0.60 m is placed in a 0.20-T uniform \vec{B} field that points toward the top of the page. Determine the magnetic flux through the ring’s area when (a) the plane of the ring is perpendicular to the surface of the page and its normal vector points to the right and (b) the plane of the ring is perpendicular to the surface of the page and its normal vector points toward the top of the page.
Answer: (a) 0; (b) 0.23 T ⋅ m².