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daser333 [38]
3 years ago
15

. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

ed at an angle of   25 with respect to the normal to the surface. What is the magnetic flux through the surface?
Physics
1 answer:
Dmitriy789 [7]3 years ago
6 0

Answer:

The magnetic flux through surface is 2.22 \times 10^{-3} Wb

Explanation:

Given :

Magnitude of magnetic field B = 0.078 T

Radius of circle r = 0.10 m

Angle between field and surface normal \theta = 25°

From the formula of flux,

\phi = B.A

\phi = BA\cos \theta

Where \theta = angle between magnetic field line and surface normal, A = area of circular surface.

A = \pi r^{2}

A = 3.14 \times (0.10) ^{2}

A = 0.0314 m^{2}

Magnetic flux is given by,

\phi = 0.078 \times 0.0314 \times \cos 25

\phi = 2.22 \times 10^{-3} Wb

Therefore, the magnetic flux through surface is 2.22 \times 10^{-3} Wb

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Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

D = \sqrt{(d_{m})^2+(d_{w})^2}

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