Homework Statement
3. A particle with magnetic dipole moment of magnitude μ (be careful—this is not a magnetic
permeability) and spin angular momentum of magnitude S is immersed in a magnetic field
of magnitude B. For simplicity, assume that the spin and magnetic-moment vectors (which
are...
Homework Statement
What are the units for electric energy density and magnetic energy density? Is it J/m^3
Homework Equations
μ(mag)= (B2)/ (2μo)
μ(electric) = (1/2) (εo) (E2)
The Attempt at a Solution
Magnetic energy density : T2/ N/A2 = J/m^3
electric energy density...
The entire question is stated as so:
Two particles, each with electric charge q and mass m, are traveling in vacuum side-by-side,
on parallel trajectories a distance d apart, both with speed v (much less than the speed of
light). Calculate the ratio of the magnitude of the magnetic force to...
Homework Statement
There are two particles side by side in a vacuum with electric charge q and mass m, traveling at distance d apart, with v speed(less than speed of light). What is the ratio of magnitude of magnetic to electric force the particle exert on each other?
Homework Equations...
well F=ma and E=F/q so E= ma/q
|Felectric| is related to E by E=F/q
and |Fgravity| related to g by F=ma
correct?
im not understanding why you have Felectric/ Fgravity=1
Vela. I'm really confused I've been working on this for hours. I know that the gravitational and electric force have to cancel in order for the particle to be hovering.how would another force be included in the equation q/m=a/E ? or am i using the wrong equation?
Okay Thanks! I found acceleration can you reassure my answer?
so F=ma and E=F/q I substitute and get a= Eq/m and with the numbers plugged in I get:
a= (-150N/C)(1.602*10^-19 Nm^2/C^2) / (1.67*10^-27 kg)
a= -1.44*10^10 m/s^2
for charge to mass ratio I used E=ma/q rearrange to get q/m= a/E...
okay I fixed the units and decimal points on my paper. it would be -0.0654 C/kg .
okay if that is correct how would I find the total energy associated with the electric field of the Earth, if Earth is a charged capacitor? Would I use U=1/2 CV^2. Does the value of capacitance of the Earth matter?
ma=Eq I divided by m on each side to get q/m= aE. Then a=-9.81 and E= -150. I multiplied. Oh wait okay so I was supposed to get: a/E= q/m. so It would be -9.81/-150 = -0.654. q/m = -0.654 Correct?