All categories

3 years ago

If the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave

Physics

1 answer:

tangare [24]3 years ago

4 0

CHECK THE COMPLETE QUESTION BELOW;

the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave? (c=3.0×108m/s, μ0=4π×10−7T⋅m/A, ε0=8.85×10−12C2/N⋅m2)

Answer:

2.4×10^5 N/C

Explanation:

the amplitude can be explained as the maximum field strength of the electric and magnetic fields. Wave energy is proportional to its amplitude squared (E2 or B2).

We were told to calculate the amplitude of the magnetic field, which can be done using expresion below

S=ε²/2cμ

Where S is the intensity intensity of an electromagnetic wave given as 80 MW/m2

ε² is the Amplitude which we are looking for

c= speed of light given as 3×10^8m/s

Substitute the values into above formula we have,

S=ε²/2cμ

Making Amplitude subject of formula

ε²=S×2cμ

ε²=[80×10^6)(2×3×10^8)(4Π×10^-7)

= 245598.44

ε²=2.4×10^5 N/C

Therefore, amplitude of the magnetic field of this wave is S=2.4×10^5 N/C

You might be interested in

chris and molly wins $2400 in a competition they share the money in the ratio2:3 how much money do they each receive ?

nlexa [21]

Answer:

chris gets 960 and molly gets 1440

Explanation:

add the ratio up and divide

2+3=5

2400/5=480

480x2=960

480x3=1440

960+1440= 2400

5 0

3 years ago

A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image

lilavasa [31]

Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor. T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

6 0

3 years ago

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$