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3 years ago

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A nonuniform linear charge distribution given by λ = bx, where b is a constant, is located along an x axis from x = 0 to x = 0.2

0 m. If b = 20 nC/m² and V = 0 at infinity, what is the electric potential at (a) the origin and (b) the point y = 0.15 m on the y axis?

1 answer:

Alexeev081 [22]3 years ago

7 0

Answer:

(a) V = 36 v

(b) V = 18v

Explanation:

Check the attached file for the explanation

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If a bar of copper is brought near a magnet, the copper bar will be

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It will be unaffected by the magnet because it has no magnetic field. If you were to maybe have electricity going through it is the only way it would have anything to do with the magnet.
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Transmission Lines and Health. Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the elec

Blababa [14]

Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:

B = μ₀I/(2πr)

B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from line

Given values:

μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m

Plug in and solve for B:

B = 4π×10⁻⁷(170)/(2π(8.0))

B = 4.25×10⁻⁶T

The earth's magnetic field strength is 0.50G or 5.0×10⁻⁵T. Calculate the ratio of the line's magnetic field strength to earth's magnetic field strength:

4.25×10⁻⁶/(5.0×10⁻⁵)

= 0.085

= 8.5%

The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.

8 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

3 years ago

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0

3 years ago

An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.

adoni [48]

Answer :

The least possible uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi}$ ...........(1)

where,

$\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

$p=m\times v$

or,

$\Delta p=m\times \Delta v$ .......(2)

Equating 1 and 2, we get:

$\Delta x\times m\times \Delta v=\frac{h}{4\pi}$

$\Delta v=\frac{h}{4\pi \Delta x\times m}$

Given:

m = mass of electron = $9.11\times 10^{-31}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

radius of atom = $67.0pm=67.0\times 10^{-12}m$ $(1pm=10^{-12}m)$

$\Delta x$ = diameter of atom = $2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}$

$\Delta v=4.32\times 10^{5}m/s$

The minimum uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

Now we have to calculate the percentage of the average speed.

Percentage of average speed = $\frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100$

Uncertainty in speed = $4.32\times 10^{5}m/s$

Average speed = $1.3\times 10^{6}m/s$

Percentage of average speed = $\frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100$

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %

3 0

3 years ago