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Hitman42 [59]
3 years ago
5

A straight wire of length 4.5 cm moves at a constant speed of 5.2m/s perpendicular to its length and a uniform magnetic field. I

f the induced e.m.f. in a straight wire is 85mV find the magnetic flux density.
Physics
1 answer:
liraira [26]3 years ago
3 0
Well you have to minus the 4.5 to 5.2 and the answer to that would be -11.5 and calculated that to be 4.5
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4 0
2 years ago
In an electricity demonstration at the Deutsches Museum in Munich, Germany, a person sits inside a metal sphere of radius 0.90 m
vladimir2022 [97]

Answer:

 E_interior = 0

Explanation:

As the sphere is metallic, the electrical charges are distributed on its surface, as far away as possible from each other.

If we apply Gauss's law, as the charge is on the surface, when drawing a spherical Gaussian surface, we see that there is no charge inside, therefore there is no electric field inside the metallic sphere.

          E_interior = 0

6 0
3 years ago
Olivia put a glass of water in the freezer. She left it there for three hours. When she returned, the water had turned to ice. W
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A. freezing, when water turns to ice the water is turning from a liquid to a solid.
7 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
Please answer this fast
Natali5045456 [20]

the answer choice will be A because they travel at the same speed through only light not one material.

3 0
3 years ago
Read 2 more answers
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