Chapter 32, Problem 24SP

A room has its walls aligned accurately with respect to north, south, east, and west. The north wall has an area of 15 ${\text{m}}^{\text{2}}$, the east wall has an area of 12 ${\text{m}}^{\text{2}}$, and the floor’s area is 35 ${\text{m}}^{\text{2}}$. At the site the Earth’s magnetic field has a value of 0.60 G and is directed 50° below the horizontal and 7.0° east of north. Find the magnetic flux through the north wall, the east wall, and the floor.

To determine

The magnetic flux through the north wall, the east wall, and the floor of the room that align accurately with respect to north, south, east, and west if the area of north wall, the east wall, and the floor is $15{\text{ m}}^2$, $12{\text{ m}}^2$, and $35{\text{ m}}^2$, respectively. The Earth’s magnetic field is $0.60\text{G}$ and is directed $50°$ below the horizontal and $7.0°$ east of north.

Answer to Problem 24SP

Solution:

$0.57\text{ mWb}$, $\text{56 μWb}$, and $\text{1}\text{.6 mWb}$.

Explanation of Solution

Given data:

The area of the north wall is $15{\text{ m}}^2$.

The area of the east wall is $12{\text{ m}}^2$.

The area of the floor is ${\text{35 m}}^2$.

The Earth’s magnetic field is $0.60\text{}\text{ G}$.

The direction of the Earth’s magnetic field is $50°$ below the horizontal line and $7.0°$ east of north.

Understand that $1\text{ G}={10}^{-4}\text{ T}$.

Formula used:

The expression for flux in terms of magnetic field and area is expressed as

$\begin{array}{c}\Phi =\overrightarrow{B}\cdot \overrightarrow{A}\\ =BA\cos \theta \end{array}$

Here, $B$ is the magnetic field, $A$ is the area, and $\theta$ is the angle between the area normal and magnetic field.

Write the expression for the area of a rectangle:

$A=lb$

Here, $l$ is the length and $b$ is the breath of the rectangle.

Explanation:

Sketch the approximate diagram of the Earth’s magnetic field and the north wall:

In figure (1), ${\theta }_{NH}$ is the east–north angle, $B_E$ is the Earth’s magnetic field, and ${\theta }_H$ is the angle between the Earth’s magnetic field and the horizontal line.

Recall the expression of flux through the north wall:

${\Phi }_{\text{North}\text{}\text{ wall}}=B_E{A}_N\cos {\theta }_H$

Here, $A_N$ is the area of the north wall.

Substitute $50°$ for ${\theta }_H$, $15{\text{ m}}^2$ for $A_N$, and $0.60\text{}\text{ G}$ for $B_E$

$\begin{array}{c}{\Phi }_{\text{North}\text{}\text{ wall}}=\left(0.60\text{}\text{ G}\right)\left(15{\text{ m}}^2\right)\cos \left(50°\right)\\ =\left(0.60\text{}\text{ G}\right)\left(15{\text{ m}}^2\right)\cos \left(50°\right)\\ =\left(0.60\text{}\text{ G}\left(\frac{{10}^{-4}\text{}\text{ T}}{1\text{ G}}\right)\right)\left(15{\text{ m}}^2\right)\cos \left(50°\right)\\ =5.78\times {10}^{-4}\text{Wb}\end{array}$

The Earth’s magnetic field is directed $7.0°$ east of north. Hence, the total magnetic flux is calculated as

${\Phi }_{\text{Total North}\text{}\text{ wall}}=\left({\Phi }_{\text{North}\text{}\text{ wall}}\right)\cos 7°$

Substitute $5.78\times {10}^{-4}\text{Wb}$ for ${\Phi }_{\text{North}\text{}\text{ wall}}$

$\begin{array}{c}{\Phi }_{\text{Total North}\text{}\text{ wall}}=\left(5.78\times {10}^{-4}\text{Wb}\right)\cos 7°\\ =0.57\text{ mWb}\end{array}$

Hence, the total flux that passes through the north wall is $0.57\text{ mWb}$.

Sketch the approximate diagram of the Earth’s magnetic field and the east wall:

Recall the expression of flux through the east wall:

${\Phi }_{\text{East}\text{}\text{ wall}}=B_E{A}_E\cos {\theta }_H$

Here, $B_E$ is the Earth’s magnetic field, $A_E$ is the area of north wall, and ${\theta }_H$ is the angle between Earth’s magnetic field and the horizontal line.

Substitute $50°$ for ${\theta }_H$, $12{\text{ m}}^2$ for $A_E$, and $0.60\text{}\text{ G}$ for $B_E$

$\begin{array}{c}{\Phi }_{\text{East}\text{}\text{ wall}}=\left(0.60\text{}\text{ G}\right)\left(12{\text{ m}}^2\right)\cos \left(50°\right)\\ =\left(0.60\text{}\text{ G}\right)\left(12{\text{ m}}^2\right)\cos \left(50°\right)\\ =\left(0.60\text{}\text{ G}\left(\frac{{10}^{-4}\text{}\text{ T}}{1\text{ G}}\right)\right)\left(12{\text{ m}}^2\right)\cos \left(50°\right)\\ =4.63\times {10}^{-4}\text{Wb}\end{array}$

The Earth’s magnetic field is directed $7.0°$ east of north. Hence, the total magnetic flux that passes through the east wall is calculated as

${\Phi }_{\text{Total East}\text{}\text{ wall}}=\left({\Phi }_{\text{East}\text{}\text{ wall}}\right)\sin 7°$

Substitute $4.63\times {10}^{-4}\text{Wb}$ for ${\Phi }_{\text{East}\text{}\text{ wall}}$

$\begin{array}{c}{\Phi }_{\text{Total East}\text{}\text{ wall}}=\left(4.63\times {10}^{-4}\text{Wb}\right)\sin 7°\\ =56\text{ μWb}\end{array}$

Hence, the total flux that passes through the east wall is $56\text{ μWb}$.

Sketch the approximate diagram of the Earth’s magnetic field and the floor:

Recall the expression of flux through the floor:

${\Phi }_{\text{Floor}}=B_E{A}_F\cos {\theta }_H$

Here, $B_E$ is the Earth’s magnetic field, $A_F$ is the area of floor, and ${\theta }_H$ is the angle between Earth’s magnetic field and the horizontal line.

Understand that the flux through the floor would be vertical component passes through the floor only. Hence, the angle is $90°-50°=40°$.

Substitute $40°$ for ${\theta }_H$, ${\text{35 m}}^2$ for $A_F$, and $0.60\text{}\text{ G}$ for $B_E$

$\begin{array}{c}{\Phi }_{\text{Floor}}=\left(0.60\text{}\text{ G}\right)\left({\text{35 m}}^2\right)\cos \left(40°\right)\\ =\left(0.60\text{}\text{ G}\left(\frac{{10}^{-4}\text{}\text{ T}}{1\text{ G}}\right)\right)\left({\text{35 m}}^2\right)\cos \left(40°\right)\\ =1.6\text{mWb}\end{array}$

Hence, the total flux that passes through the floor is $1.6\text{mWb}$.

Conclusion:

Therefore, the total flux that passes through the north wall is $0.57\text{ mWb}$, the total flux that passes through the east wall is $56\text{ μWb}$, and the total flux that passes through the floor is $1.6\text{mWb}$.