Question

A house has a floor area of 112m², and has an outside wall that has an area of 28m². The earth’s magnetic field here has a horizontal component of $2.6\times {10}^{-5}T$ that points due north and a vertical component of $4.2\times {10}^{-5}T$ that points straight down, toward the earth. Determine the magnetic flux through the wall if the wall faces (a) north, and (b) east, and (c) also calculate the magnetic flux that passes through the floor.

Answer 1

Step 1: Given the data and formula used

Floor area, $A_F=112m^2$

Outside wall area, $A_W=28m^2$

The horizontal component of the earth’s magnetic field, $B_H=2.6\times {10}^{-5}T$

The vertical component of earth’s magnetic field, $B_H=4.2\times {10}^{-5}T$

Magnetic flux crossing through a surface is $\varphi =\overrightarrow{B}.\overrightarrow{A}=BA\cos \theta$

Step 2: Finding the magnetic flux

(a) Magnetic flux through the wall facing north, $\varphi =B_H{A}_W\cos \theta$

$=2.6\times {10}^{-5}T\times 28m^2\times \cos 0^o=2.6\times {10}^{-5}T\times 28m^2\times 1$

$=72.8Wb$

Magnetic flux through the wall facing north is 72.8 Wb

(b) As the field is in the perpendicular direction to the east, magnetic flux through the wall facing east, $\varphi =B_H{A}_W\cos \theta$

$=2.6\times {10}^{-5}T\times 28m^2\times \cos {90}^o=2.6\times {10}^{-5}T\times 28m^2\times 0$

$=0$

Magnetic flux through the wall facing north is 0

(c) Magnetic flux that passes through the floor, $\varphi =B_V{A}_F\cos \theta$

$=4.2\times {10}^{-5}T\times 112m^2\times \cos 0^o=4.2\times {10}^{-5}T\times 112m^2\times 1$

$=470.4Wb$

The magnetic flux that passes through the floor is 470.4Wb.

Step 3: Final answer

Thus the magnetic flux through the wall if the wall faces (a) north is 72.8Wb, (b) east is 0, and (c) the magnetic flux that passes through the floor is 470.4Wb.