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3 years ago

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Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor

m magnetic field that is directed at an angle of 31.0 ∘ above the horizontal. Part A What must the magnitude of the magnetic field be in order to produce a flux of 4.00×10−4 Wb through the surface? Express your answer with the appropriate units. BB = nothing nothing Request Answer Provide Feedback

1 answer:

deff fn [24]3 years ago

7 0

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, $A=8.82\ cm^2=0.000882\ m^2$

Angle between the uniform magnetic field and the horizontal, $\theta=31$

Magnetic flux, $\phi=4\times 10^{-4}\ Wb$

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

$\phi=BA\ cos\theta$

$\theta$ is the angle between magnetic field and the area

Here, $\theta=90-31=59^{\circ}$

$B=\dfrac{\phi}{A\ cos\theta}$

$B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}$

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

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