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2 years ago

At what rate must electrons in a wire vibrate to emit microwaves with a wavelength of 1.00 mm

2 answers:

7 0

The equation to use is v=fλ where v is the speed of light, f is the frequency, i.e. the rate at which they vibrate, and λ is the wavelength in meters.

Rearrange the above equation and you get f=v/λ. So the answer is f=300000000/0.001 or 300GHz.

AnnyKZ [126]2 years ago

5 0

Answer:

f = 3×10¹¹ Hz

Explanation:

Relation ship between frequency and wavelength

"The wave speed (v) is defined as the distance traveled by a wave per unit time. If considered that the wave travels a distance of one wavelength in one period,

ν=λ/T

As we know that T = 1/f, hence we can express the above equation as,

V = f λ

The wave speed is equal to the product of its frequency and wavelength, and this implies the relationship between frequency and wavelength."

The relation between frequency and wavelength is

λ×f = c

c = speed of light = 3×10⁸ m/s

λ = 1.00 mm = 10⁻³ m

f=c/λ

f=(3×10⁸ m/s)/ 10⁻³ m

f = 3×10¹¹ Hz

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The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

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2 years ago

he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th

AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x $10^{-6}$ T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ $_{0}$ I/2πr

where

B = magnetic field (T)

μ $_{0}$ = 1.26 x $10^{-6}$ (N/ $A^{2}$ )

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.19)

= 4.96 x $10^{-6}$ T

BV = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.41)

= 2.3 x $10^{-6}$ T

hence,

the net magnetic field = BH - BV

= 4.96 x $10^{-6}$ - 2.3 x $10^{-6}$

= 2.66 x $10^{-6}$ T

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1 year ago