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3 years ago

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A beam of protons enter the electric field of magnitude E = 0.5 N/C between a pair of parallel plates. There is a magnetic field

between the plates. The magnetic field is parallel to the plates and perpendicular to the initial direction of the protons, and its magnitude is 2.3 T. The proton beam passes through undeflected. What is the speed of the protons?

1 answer:

HACTEHA [7]3 years ago

3 0

Answer:

0.217 m/s

Explanation:

The protons in the beam passes undeflected when the electric force is equal to the magnetic force:

qE = qvB

where

q is the proton's charge

E is the magnitude of the electric field

v is the speed of the protons

B is the magnitude of the magnetic field

Re-arranging the equation,

$v=\frac{E}{B}$

And by substituting

E = 0.5 N/C

B = 2.3 T

We find

$v=\frac{0.5}{2.3}=0.217 m/s$

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The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

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Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

$\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma = 6.88 \times 10^{14}\ s^{-1}$

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2 years ago

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Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

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where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

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Answer:

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