A $2.00$-m long metal wire is formed into a square and placed in the horizontal $x y$-plane. A uniform magnetic field is oriented at $30 ^ { \circ }$ above the horizontal with a strength of $0.344 \mathrm {~T}$. What is the magnetic flux through the square due to this field?
A) $0.0745 \mathrm {~T} \cdot \mathrm { m } ^ { 2 }$
B) $0.0430 \mathrm {~T} \cdot \mathrm { m } ^ { 2 }$
C) $0.172 \mathrm {~T} \cdot \mathrm { m } ^ { 2 }$
D) $0.298 \mathrm {~T} \cdot \mathrm { m } ^ { 2 }$

Question

Quizplus · Accepted Answer

The area of the square is $(2.00 \, \text{m} / 4)^2 = 0.25 \, \text{m}^2$. The magnetic flux, $\Phi = B A \cos(\theta)$, where $B = 0.344 \, \text{T}$, $A = 0.25 \, \text{m}^2$, and $\theta = 30^\circ$. Thus, $\Phi = 0.344 \times 0.25 \times \cos(30^\circ) = 0.0430 \, \text{T} \cdot \text{m}^2$.

A 2.002.002.00 -M Long Metal Wire Is Formed into a Square and Square

A $2.00$ -M Long Metal Wire Is Formed into a Square and Square