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tamaranim1 [39]
2 years ago
3

Determine the magnitude of the magnetic flux through the south-facing window of a house in British Columbia, where Earth's \over

rightarrow{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.
Physics
1 answer:
77julia77 [94]2 years ago
4 0

Answer:

0

Explanation:

B = Magnetic field = 5.8\times 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\times 10^{-5}\times 4.5\times \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

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Answer:

The electric field strength is 4.5\times 10^{4} N/C

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7 0
2 years ago
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