Question

Determine the magnitude of the magnetic flux through the south-facing window of a house in British Columbia, where Earth's \overrightarrow{B} field has a magnitude of 5.8 × 10?5 T and the direction of \overrightarrow{B} field is 72? below horizontal with the horizontal component of\overrightarrow{B}field directed to the north. Assume the area of the window is 4.5 m^{2}

Question

Determine the magnitude of the magnetic flux through the west-facing window that has the same area.

Answer 1

0

Step-by-step explanation:

B = Magnetic field =
`5.8* 10^(-5)\ T`

A = Area of window =
`4.5\ m^2`

`\theta` =Angle of the magnetic field =
`90^(\circ)`

Magnetic flux is given by

`\phi=BA\cos \theta\\\Rightarrow \phi=5.8* 10^(-5)* 4.5* \cos 90^(\circ)\\\Rightarrow \phi=0\ Tm^2`

The magnitude of the magnetic flux through the west-facing window that has the same area is 0