Radiation Pressure: Solving for Magnetic Field

AI Thread Summary
The discussion revolves around a physics homework problem involving radiation pressure used to suspend a piece of paper. The paper's area and mass are provided, and the student initially struggles with calculating the required light power and determining the magnetic field's magnitude. They realize that the pressure can be calculated using the force of gravity acting on the paper, leading to the formula u = F/A. After some contemplation, the student concludes they have solved the problem correctly by relating the pressure to the weight of the paper. The conversation highlights the importance of understanding the relationship between force, area, and radiation pressure in physics.
jarfungus
Messages
2
Reaction score
0
Hello! First time on this site, so I hope I do this right. I have a homework question that I could use some help on:

(From Physics for Scientists and Engineers, Third Ed. Fishbane, Gasiorowicz, Thornton)

Chapter 34 #39.
Suppose that you want to use the radiation pressure from a beam of light to suspend a piece of paper in a horizontal position; the paper has an area of 50 cm^2 and a mass of 0.20 g. Assume that their is no problem with the balance, that the paper is dark and absorbs the beam fully, and that the entire beam can be used to hold the paper against the pull of gravity. How many watts must the light produce? Given your answer, what do you think will happen to the paper?

Since power=c*u*A, I can easily calculate the amount, right? The problem comes when solving for u. I know that u=1/u0*B^2. But how can I find the magnitude of the magnetic field with the information given? Would it be right to manipulate the formulas so that B=E^2/(sqrt(1-c^2))? (Then what is E?) ...



 
Physics news on Phys.org
I got it:)

Nevermind! I think I figured it out. I forgot the u=F/A, and the force is m*g...so I am all set:)
Thanks for all the viewing.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top