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77julia77 [94]
3 years ago
9

A current I flows down a wire of radius a.

Physics
1 answer:
Helga [31]3 years ago
7 0

Answer:

(a) K = \frac{I}{2\pi a}

(b) J = \frac{I}{2\pi as}

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

                                   K = \frac{dI}{dL}

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

                                   dL = 2\pi r

The radius of the wire = a

                                    dL = 2\pi a

The surface current density K = \frac{I}{2\pi a}

(b) The current density is inversely proportional

                                     J \alpha  s^{-1}    

                                     J = \frac{k}{s}           ......(1)

k is the constant of proportionality

                                     I = \int\limits {J} \, dS

                                     I = J \int\limits \, dS     ........(2)

substituting (1) into (2)

                                     I = \frac{k}{s} \int\limits\, dS

                                     I = k \int\limits^a_0 \frac{1}{s}  {s} \, dS

                                     I = 2\pi k\int\limits\, dS

                                     I = 2\pi ka

                                     k = \frac{I}{2\pi a}

substitute J = \frac{k}{s}

                                     J = \frac{I}{2\pi as}

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A balloon that has been rubbed on wool has a negative charge. When the balloon is held over—but not touching—some bits of paper,
alexandr402 [8]

Answer:

This process is known as static electricity

Explanation:

When balloon is rubbed on the wool , a negative charge is usually formed. When the balloon is then held over—but not touching—some bits of paper, some of the bits jump up to the balloon. The bits which jump up to the balloon are positive charged and we know that unlike charges attract.

This process explains static electricity which is when the charges on a material are not in a balanced state.

5 0
3 years ago
2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela
Vanyuwa [196]

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

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3 years ago
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Answer:

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3 years ago
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A boy standing on a bridge above a river throws stone A vertically upward with an initial velocity =15m/s.2 seconds later he dro
Anastaziya [24]

I'm not sure if this is correct but it's what I'll do

This is free-fall problem.
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Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )

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Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2

4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t

Time for Stone B is 4s
Time for Stone A is 6s

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How do sinkholes occur? Overburden falls into the space left behind when bedrock is dissolved. Bedrock falls under the pressure
BabaBlast [244]

Answer: over burden is dissolved by water wind and acids

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