COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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Chapter 22, Problem 36QAP
To determine

(a)

The amplitude of the electric field.

Expert Solution
Check Mark

Answer to Problem 36QAP

The amplitude of the magnetic field is given by,

Bo=0.7μT

Explanation of Solution

Given:

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]

Formula used:

The magnetic field is given,

B(x,t)=Bocos[kxωt+ϕ]

Where,

B(x,t) =Magnetic field Bo = Amplitude of magnetic field k = Wave number

ω =Angular frequency

ϕ =Phase difference

Calculation:

The magnetic field is given,

B(x,t)=Bocos[kxωt+ϕ]...(1)

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]B(x,t)=(0.7μT)cos[(8π×106m1)x(2.40π×1015s1)t+ϕ]...(2)

Comparing equation (1) and (2) we get,

Bo=0.7μT

Thus, the amplitude of the magnetic field is given by,

Bo=0.7μT

To determine

(b)

The speed of the electromagnetic wave.

Expert Solution
Check Mark

Answer to Problem 36QAP

The speed of the electromagnetic wave is, c=3×108m/s

Explanation of Solution

Given:

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]

Formula used:

The speed of electromagnetic wave is is,

c=ωk

Where,

c =Speed of electromagnetic wavek = Wave number

ω =Angular frequency

Calculation:

The magnetic field is given,

B(x,t)=Bocos[kxωt+ϕ]...(1)

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]B(x,t)=(0.7μT)cos[(8π×106m1)x(2.40π×1015s1)t+ϕ]...(2)

Comparing equation (1) and (2) we get,

Bo=0.7μTω=2.40π×1015s1k=8π×106m1

Thus, the speed of the magnetic field is given by,

c=ωkc=2.40π×10158π×106c=3×108m/s

To determine

(c)

The frequency of the electromagnetic wave.

Expert Solution
Check Mark

Answer to Problem 36QAP

The frequency of the electromagnetic wave is, f=1.200×1014Hz

Explanation of Solution

Given:

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]

Formula used:

The angular frequency of electromagnetic wave is is,

ω=2πf

Where,

ω =Angular frequency

f =Frequency

Calculation:

The magnetic field is given,

B(x,t)=Bocos[kxωt+ϕ]...(1)

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]B(x,t)=(0.7μT)cos[(8π×106m1)x(2.40π×1015s1)t+ϕ]...(2)

Comparing equation (1) and (2) we get,

Bo=0.7μTω=2.40π×1015s1k=8π×106m1

The angular frequency of electromagnetic wave is is,

ω=2πf2.40π×1015=2πff=2.40π×10152πf=1.200×1014Hz

To determine

(d)

The period of the electromagnetic wave.

Expert Solution
Check Mark

Answer to Problem 36QAP

The period of the electromagnetic wave is, T=8.33×1015s

Explanation of Solution

Given:

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]

Frequency, f=1.200×1014Hz

Formula used:

The period of electromagnetic wave is,

T=1f

Where,

T =Period

f =Frequency

Calculation:

The period of electromagnetic wave is,

T=1fT=11.200×1014T=8.33×1015s

To determine

(e)

The wavelength of the electromagnetic wave.

Expert Solution
Check Mark

Answer to Problem 36QAP

The wavelength of the electromagnetic wave is, λ=0.25×106m

Explanation of Solution

Given:

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]

Formula used:

The wave number of electromagnetic waves is,

k=2πλ

Where,

k =Wave-number

λ =Wavelength

Calculation:

The magnetic field is given,

B(x,t)=Bocos[kxωt+ϕ]...(1)

B(x,t)=(0.7μT)sin[(8π×106m1)x(2.40π×1015s1)t]B(x,t)=(0.7μT)cos[(8π×106m1)x(2.40π×1015s1)t+ϕ]...(2)

Comparing equation (1) and (2) we get,

Bo=0.7μTω=2.40π×1015s1k=8π×106m1

The wave number of electromagnetic waves is,

k=2πλ8π×106=2πλλ=2π8π×106λ=0.25×106m

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