Electric fields, Pulses, Electromagnetic energy

[KU_Mustang]

I'm working on these problems can anyone help me with it?

1. In an EM wave traveling west, the B field oscillates vertically and has a frequency of 56.0 kHz and an rms strangth of 7.90E-9 T. What is the rms strength of the electric field?


2. Pulsed lasers used for science and medicine produce very short bursts of electromagnetic energy. The laser light wavelength is 1.040 E3 nm, and the pulse lasts for 55 picoseconds. How many wavelengths are found within the laser points? How short would the pulse need to be to fit only one wavelength?


3. A mirror at an amusement park shows an upright image of any person who stands 1.3 m in front of it. If the image is four times the person's height, what is the radius of the curvature?


4. The image of a distant tree is virtual and very small when viewed in a curved mirror. The image appears to be 15.0 cm behind the mirror. The mirror is convex. What is the radius of the curvature?

 

[KU_Mustang]

Nevermind, thanks :) I already have them answered!

 

[thepatientmental]

1. To find the rms strength of the electric field, we can use the relationship between the electric and magnetic fields in an EM wave:

E/B = c

Where c is the speed of light in a vacuum. Plugging in the given values, we get:

E/7.90E-9 T = 2.99E8 m/s

Solving for E, we get an rms strength of 2.36E-1 V/m.

2. To find the number of wavelengths in the pulse, we can use the formula:

Number of wavelengths = pulse duration/wavelength

Plugging in the given values, we get:

Number of wavelengths = (55 picoseconds)/(1.040 E3 nm)

= 5.29E-8/1.040E3

= 5.08E-14 wavelengths

To fit only one wavelength, the pulse would need to be 1.040 E3 nm or shorter.

3. The radius of curvature of a mirror is related to the distance between the object and the mirror, and the height of the object and its image:

1/f = 1/di + 1/do

Where f is the focal length, di is the distance of the image from the mirror, and do is the distance of the object from the mirror.

Plugging in the given values, we get:

1/f = 1/1.3 + 1/4.0

Solving for f, we get a radius of curvature of 1.69 m.

4. In a convex mirror, the radius of curvature is twice the focal length:

f = 2R

Where f is the focal length and R is the radius of curvature.

Plugging in the given values, we get:

f = 2(15.0 cm)

= 30.0 cm

Therefore, the radius of curvature is 30.0 cm.