Question

Determine the magnitude of the magnetic flux through the south-facing window of a house in British Columbia, where Earth's B⃗ field has a magnitude of 5.8 × 10−5 T and the direction of B⃗ field is 72∘ below horizontal with the horizontal component of B⃗ field directed to the north. Assume the area of the window is 3.5 m2 .

Answer 1

The following equation represents the flux of a magnetic field through a surface:

Φ = ∫∫B•dA

B is the magnetic field vector

dA is the vector normal to the surface element

We are integrating the dot product of B and dA over the area of the window.

B and dA is constant everywhere, therefore we can simplify the calculation to the following equation:

Φ = BAcos(θ)

B is the magnetic field strength

A is the area of the window

θ is the angle between the magnetic field and the window's normal vector

Our perspective is such that left and right orientations represent north and south. The window faces south, so its normal vector faces horizontally. The earth's magnetic field is oriented 72° below the horizontal, therefore θ = 72°

Given values:

B = 5.8×10⁻⁵T

A = 3.5m²

θ = 72°

Plug in the values and solve for Φ:

Φ = (5.8×10⁻⁵)(3.5)cos(72°)

Φ = 6.3×10⁻⁵Wb

Answer 2

The magnitude of the magnetic flux $\phi=6.3\times10^{-5}\ Wb$

What will be the Magnetic flux through a window?

The equation for the magnetic flux is given by

$\phi=\int\limitsa { }\int\limitsa {B} \, dA$

B is the magnetic field vector

dA is the vector normal to the surface element

We are integrating the dot product of B and dA over the area of the window.

By solving the following equation

$\phi =BAcos(\theta)$

B is the magnetic field strength

A is the area of the window

θ is the angle between the magnetic field and the window's normal vector

Our perspective is such that left and right orientations represent north and south.

The window faces south, so its normal vector faces horizontally.

The earth's magnetic field is oriented 72° below the horizontal, therefore θ = 72°

The given values are

$B= 5.8\times 10^{-5}T$

$A=3.5m^2$

$\theta =72$

Put the values

$\phi= (5.8\times10^-{5}(3.5)cos(72)$

$\phi=6.3\times10^{-5}\ Wb$

Thus the magnitude of the magnetic flux $\phi=6.3\times10^{-5}\ Wb$

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