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Kobotan [32]
3 years ago
11

A cylindrical conductor with a circular cross section has a radius a and a resistivity p and carries a constant current I. (Take

the current to be coming out of the page when the cross-sectional view of the conductor is in the plane of the page.) For the following questions, express your answer in terms of the variables I,a,p, and appropriate constants (u0,e0, and pi).a) what is the magnitude of the electric field vector E at a point just inside the wire at a distance a from the axis?b) What is the magnitude of the magnetic field vector B at the same point?c) What is the magnitude of the Poynting vector S at the same point?d) What are the directions of these three vectors?
Physics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

Explanation:

a)

Using Ohms Law

R= \rho \frac{l}{\pi a^2 }
\\V = IR = E l = I \rho \frac{l}{\pi a^2 }
\\E = \frac{I \rho}{\pi a^2 }
\\E = J \rho


Where J is the current density J = \frac{I}{\pi a^2 }


and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor E = \rhoJ just inside at a radius a (and anywhere else).

b)

Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current

\oint B .dl = B 2 \pi a = \mu_{o} I


such that

B = \frac{\mu_{o} I}{2 \pi a }


and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.

c)

The Poynting vector is given by

S = \frac{1}{ \mu_o} |E \textrm{x}B| = \frac{\rho I^2}{2 \pi^2 a^3}


and by the right hand rule it’s always pointing in towards the center of the conductor.

d)

Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c

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Colt1911 [192]

Answer:

(A) <u>The same</u>

(B) <u>More</u>

(C) The magnitude of the parent's acceleration is 0.625 m/s²

Explanation:

The given parameters are;

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The mass of the child, m₂ = 16 kg

(A) By Newton's third law of motion, action and reaction are equal and opposite

Therefore, the action of the parent on the child is equal to the reaction of the child on the parent and vice versa

Therefore, the force experienced by the child is <u>the same</u> as the force experienced by the parent

(B) Newton's second law states that an objects acceleration is directly proportional to the applied force and inversely proportional to the mass of the object

Therefore, the parent and the child both experience the same force but the mass of the child is less than the mass of the parent and therefore, by Newton's second law, the acceleration of the child will be <u>more</u> than the acceleration of the parent for the same given force

(C) The acceleration of the child, a₂ = 2.5 m/s²

Let F₁ represent the force experienced by the parent, let F₂ represent the force experienced by the child and let a₁ represent the magnitude of the parent's acceleration

By Newton's third law, we have;

F₁ = F₂

Force, F = Mass, m × Acceleration, a

We can write, F = m × a

Therefore;

F₁ = m₁ × a₁ and F₂ = m₂ × a₂

∴ F₁ = F₂ gives;

m₁ × a₁ = m₂ × a₂

a₁ = (m₂ × a₂)/m₁ = (16 × 2.5)/64 = 0.625

∴ The magnitude of the parent's acceleration = a₁ = 0.625 m/s²

5 0
3 years ago
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6 0
3 years ago
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Oliga [24]

Answer: white shirt-reflect

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6 0
3 years ago
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balandron [24]
B. a combustion reaction involves a hydrocarbon burning with oxygen gas to produce water and carbon dioxide.
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3.  I am pretty sure it is referring to the activation energy which is a characteristic all chemical reaction have and in a combustion reaction the spark gets the reaction over the activation energy.
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4 0
3 years ago
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Alexus [3.1K]

Answer:

0 m/s^2

Explanation:

Acceleration is the change in velocity / time

=> In this case there isn't any change in velocity

=> It is = to 0.

7 0
3 years ago
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