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Sveta_85 [38]
3 years ago
13

Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric e

ld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
Physics
1 answer:
kvv77 [185]3 years ago
5 0

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

<em>Hint </em>: ΔV = Ed <em> </em>

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F_{net = 0

so, F_E = F_B

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

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2 years ago
A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel
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(a) 7.32\cdot 10^7 Hz

The frequency of an electromagnetic waves is given by:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of light

\lambda=4.1 m is the wavelength of the wave in the problem

Substituting into the equation, we find

f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz

(b) 4.60\cdot 10^8 rad/s

The angular frequency of a wave is given by

\omega = 2\pi f

where

f is the frequency

For this wave,

f=7.32\cdot 10^7 Hz

So the angular frequency is

\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s

(c) 1.53 m^{-1}

The angular wave number of a wave is given by

k=\frac{2\pi}{\lambda}

where

\lambda is the wavelength of the wave

For this wave, we have

\lambda=4.1 m

so the angular wave number is

k=\frac{2\pi}{4.1 m}=1.53 m^{-1}

(d) 1.03\cdot 10^{-6}T

For an electromagnetic wave,

E=cB

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

I=\frac{E^2}{2\mu_0 c}

where we have

E = 310 V/m is the amplitude of the electric field

\mu_0 is the vacuum permeability

c is the speed of light

Substituting into the formula,

I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2

(g) 1.53\cdot 10^{-8} kg m/s

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

\frac{dp}{dt}=\frac{A}{c}

where the <S> is the magnitude of the Poynting vector, given by

=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2

and where the surface is

A = 1.8 m^2

Substituting, we find

\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s

(h) 8.47\cdot 10^{-7} N/m^2

For a surface that totally absorbs the wave, the radiation pressure is given by

p=\frac{}{c}

where we have

=254.2 W/m^2

c=3\cdot 10^8 m/s

Substituting, we find

p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2

8 0
3 years ago
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