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crimeas [40]
2 years ago
11

A 0.70-m radius cylindrical region contains a uniform electric field that is parallel to the axis and is increasing at the rate

5.0 × 1012 v/m?s. the magnetic field at a point 1.2 m from the axis has a magnitude of:

Physics
2 answers:
arlik [135]2 years ago
8 0

The magnetic field at a point 1.2 m from the axis has a magnitude of 7.0 × 10^–6 T

<h3>Further explanation </h3>

Maxwell's equation is a set of coupled partial differential equations that together with the Lorentz force law form the classical electromagnetism, classical optics, and electric circuits.

Integral form in the absence of magnetic or polarizable media are Gauss' law for electricity, Gauss' law for magnetism, Faraday's law of induction, Ampere's law

A 0.70 m radius cylindrical region contains a uniform electric field that is parallel to the axis and is increasing at the rate 5.0* 10^{12} v/m.s  The magnetic field at a point 1.2 m from the axis has a magnitude of?

The Maxwell’s law of induction is as follows. Consider the charging of our circular  plate capacitor , B field also  induced at  point 2.  When capacitor stops charging  B field disappears.

By using the Maxwell’s law of induction for a circle of radius r.

2\pi rB = \epsilon_{0}\mu_{0}\pi r^2 \frac{dE}{dt} , B = \frac{1}{2}  \epsilon_{0}\mu_{0}r\frac{dE}{dt} = 7*10^{-6}T

<h3>Learn more</h3>
  1. Learn more about magnetic field brainly.com/question/1687280
  2. Learn more about parallel to the axis  brainly.com/question/1461505
  3. Learn more       about a uniform electric field brainly.com/question/13105969

<h3>Answer details</h3>

Grade: 9

Subject:  physics

Chapter:  electric field

Keywords: magnetic field, a uniform electric field, parallel,  the axis,  point

Svetllana [295]2 years ago
5 0
Magnetic fields are effects that are being produced by electric currents that are present in a material. It is characterized by both magnitude and direction so it should be a vector quantity. Magnetic field is represented by B and is proportional to the current that is passing through. It is equal to the product of a constant, uo, and the current divided by the product of 2(pi) and the distance from the wire. Base from the problem statement, there was no value of current mentioned. Therefore, the magnetic field would be zero. No magnetic fields are produced when current is zero.
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RoseWind [281]

Answer:

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Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

                x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

               x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

               R_{E} = 6.37 10⁶ m

                tex]R_{Sum}[/tex] = 6.96 10⁸ m

                Distance = 1.5 10¹¹ m

                x_Earth = 1.5 10¹¹ / 6.96 10⁸  1.5

                x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

            d_sphere = 1.5 / 6.96 10⁸  6.37 10⁶

            d_sphere = 1.37 10⁻² mm

5 0
2 years ago
Students repeat the experiment but replace block X and block Y with block W and block Z , as shown in Figure 3. Block W and bloc
Lena [83]
The block Z would be seen in figure 10 when 4 strident turn around
8 0
2 years ago
A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi
Sholpan [36]

Answer:

i = 7.83 \mu A

Explanation:

Induced EMF in the coil is given by the equation

EMF = M\frac{di}{dt}

so we have

M = 31 \mu H

also we know that rate of change in current in solenoid is given as

\frac{di}{dt} = 2.5 A/s

so induced EMF of coil is given as

EMF = (31 \times 10^{-6})(2.5)

EMF = 77.5 \times 10^{-6} A/s

now induced current in the coil will be given as

i = \frac{EMF}{R}

i = \frac{77.5 \times 10^{-6}}{9.9}

i = 7.83 \mu A

4 0
2 years ago
How much does a person weigh if it takes 700 kg*m/s to move them 10 m/s<br><br> NEED ASAP
madreJ [45]

Answer:

\huge\boxed{m = 70 \ kg}

Explanation:

<u>Given Data:</u>

Momentum = P = 700 kg m/s

Velocity = v = 10 m/s

<u>Required:</u>

Mass = m = ?

<u>Formula:</u>

P = mv

<u>Solution:</u>

m = P / v

m = 700 / 10

m = 70 kg

\rule[225]{225}{2}

Hope this helped!

<h3>~AnonymousHelper1807</h3>
5 0
2 years ago
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia
ycow [4]
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius  is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
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  = 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
       = 0.5*(0.15 kg)*(0.5 m/s)²
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The rotational KE is
KE₂ = (1/2)*I*ω²
       = 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
       = 0.0038 J

The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J


4 0
2 years ago
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