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valentinak56 [21]

3 years ago

11

In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl

e of radius 5.7 × 10-11 m. Considering the orbiting electron to be a small current loop, determine the magnetic moment associated with this motion. (Hint: The electron travels around the circle in a time equal to the period of the motion.)

1 answer:

irga5000 [103]3 years ago

5 0

Answer:

$1.5048\times 10^{-23}\ Am^2$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

r = Radius of circle = $5.7\times 10^{-11}\ m$

v = Velocity of proton = $3.3\times 10^6\ m/s$

Magnetic moment is given by

$M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2$

The magnetic moment associated with this motion is $1.5048\times 10^{-23}\ Am^2$

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Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

Answer:

For the point inside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

For the point outside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}$

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}$

We have to integrate it over the ring, which is an angular integration.

$E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}$

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

<u>x = +x0: </u>

We will only change the boundaries of the last integration.

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

6 0

3 years ago

Which has the most momentum? (look at picture please)!!

aliina [53]

Answer: the top one

Explanation: hope this helps

6 0

2 years ago

According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the

slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

$A=\dfrac{\pi}{4}\times d^2$

Put the value into the formula

$A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2$

$A=1.96\times10^{-9}\ m^2$

We need to calculate the magnitude of the force

Using formula of force

$F=\sigma A$

Put the value into the formula

$F=196\times10^{6}\times1.96\times10^{-9}$

$F=0.38416\ N$

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

$strain=\dfrac{\Delta l}{l_{0}}$

$\Delta l=strain\times l_{0}$

Put the value into the formula

$\Delta l=0.380\times l_{0}$

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

$l=l_{0}+\Delta l$

Put the value into the formula

$I=l_{0}+0.380\times l_{0}$

$l=1.38l_{0}$

$l_{0}=\dfrac{l}{1.38}$

$l_{0}=\dfrac{12\times10^{-2}}{1.38}$

$l_{0}=0.0869\ m$

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0

3 years ago

Please help asap! Thank you.

olganol [36]

Answer:

direct current

Explanation:

it has a direct path to go down to reach the specific point

4 0

3 years ago

Read 2 more answers

A bird flew 16 km west in 5 hours , then flew 20 km east in 6 hours . What was the birds velocity?

LenaWriter [7]

During that period of time, the bird's displacement was 4 km east. So its velocity was (4km east)/(11hrs). That's 0.36 km/hour east. (rounded)

5 0

2 years ago