Question

(13%) Problem 4: The magnetic field of an electromagnetic wave is described by B, = Bocos(kor - wt), where Bo = 8.5 x 10-6 T and a = 3.5 x 107 rad/s. Randomized Variables Bo= 8.5 x 10-6 T w = 3.5 x 107 rad/s >> A 20% Part (a) What is the amplitude of the corresponding electric field oscillations. Eo, in terms of Bo? E= Grade Summary Deductions 0% Potential 10046 Late Work % 80% Late Potential 80% a B 7 8 9 HOME A Во с d g ذرا TA 4 5 6 12 + - 0 END NO BACKSPACE DEL CLEAR Submissions Attempts remaining: 6 (0% per attempt) detailed view h i j m P t Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. Submission History All Date times are displayed in Pacific Standard Time. Red submission date times indicate late work Date Time Answer Hints Feedback A 20% Part (b) Calculate the electric field Eo in volts per meter. A 20% Part (c) What is the frequency of the electromagnetic wave, f, in terms of o? A 20% Part (d) What is the wavelength of the electromagnetic wave. 1, in terms of a and the speed of light c? A 20% Part (e) Calculate the wavelength 1 in meters.

Answer 1

Part A.

Relation between electric field and magnetic field is given by:

In EM waves:

Electrostatic force due to electric field = Magnetic force

Fe = Fm

q*E0 = q*c*B0

E0 = c*B0

Where c = speed of light = 3.00*10^8 m/s

Part B.

Given that B0 = 8.5*10^-6 T

So,

E0 = 3.00*10^8*8.5*10^-6

E0 = 2550 V/m = 2.55*10^3 V/m

Part C.

frequency of EM wave will be:

Angular frequency = w = 2*pi*f

f = w/(2*pi)

-- 27

Part D.

Relation between frequency and wavelength is given by:

wavelength, $\lambda$ = c/f

$\lambda$ = c/(w/(2*pi)) = 2*pi*c/w

2лс А = а

Part E.

Given that: w = 3.5*10^7 rad/sec

So,

$\lambda$ = 2*pi*3.00*10^8/(3.5*10^7) = 53.856 m

$\lambda$ = 53.9 m = wavelength

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