Asked by Annoyingmous
Two charged particles initially are traveling in the positive x direction, each with the same speed v, and enter a non-zero magnetic field at the origin O. The magnetic field for x>0 is constant, uniform and points out of the page. In the magnetic field, their trajectories both curve in the same direction, but describe semi-circles with different radii. The radius R2 of the semi-circle traced out by particle 2 is bigger than the radius R1 of the semi-circle traced out by particle 1, and R2R1=7.09. Note that the drawing is not to scale, so the ratio of radii may not be represented accurately. Assume that the particles are sent through at different times so that they do not interact with each other.
(a) Assume that the two particles have the same mass m, but have different charges, q1 and q2. What is the ratio q2q1?
(b) How much energy did the particle with charge q2 gain traversing the region with magnetic field?
(c) Assume that instead of different charges as shown in the figure, the two particles have the same charge q, but have different mass m1 and m2. What is the ratio m2m1?
(a) Assume that the two particles have the same mass m, but have different charges, q1 and q2. What is the ratio q2q1?
(b) How much energy did the particle with charge q2 gain traversing the region with magnetic field?
(c) Assume that instead of different charges as shown in the figure, the two particles have the same charge q, but have different mass m1 and m2. What is the ratio m2m1?
Answers
Answered by
Phy
b) 0
Answered by
Annoyingmous
thanks a lot phy. any idea about the rest/some others?
Answered by
Phy
No, do you have any idea for others problems?
Answered by
Phy
I got 4th problem
Answered by
Anonymous
Please Phy 4th sol?
Answered by
Phy
B=n*E_0/c
wavelength= 2*pi*c/(n*omega)
Direction -z and -y
wavelength= 2*pi*c/(n*omega)
Direction -z and -y
Answered by
Anonymous
Thanks Phy!
Answered by
P
This wave is traveling through a medium whose index of refraction is 1.3.
E= 2.8 xsin(2*π/lamdaz + 1.19e16) t V/m
B =B_0sin(2*π/lamdaz + 1.19e16 t) Tesla
Please tel ans for this phy.
E= 2.8 xsin(2*π/lamdaz + 1.19e16) t V/m
B =B_0sin(2*π/lamdaz + 1.19e16 t) Tesla
Please tel ans for this phy.
Answered by
Anonymous
2, 3, 5, 7 ,8 pls Phy
Answered by
Phy
No i didn't get them yet, I'll tell you when i get those, can u tell me the solutions of 1,4,6,9,10,11,12?
Answered by
Phy
P, these are the answers-
1.2133e-8
1.21845e-7
1.2133e-8
1.21845e-7
Answered by
dd
E⃗ = 8.6 xˆsin(2πλz + 5.91e16 t) V/m
B⃗ = Bo⃗ sin(2πλz + 5.91e16 t) Tesla
o can you explain the formula what is n, c and omega?
B⃗ = Bo⃗ sin(2πλz + 5.91e16 t) Tesla
o can you explain the formula what is n, c and omega?
Answered by
robbo
a) q2/q1 = R1/R2
b) 0
c) R2/R1
b) 0
c) R2/R1
Answered by
Jack
Phy, do you have problem 8?
Answered by
P
robbo ppls tell what are the formulas for which you have posted.
Answered by
P
Phy thanks for the ans I have ans for Q3 a=0 and last part =0
Answered by
robbo
P, just trust me and go for the answers with the solns provided. We barely have time, so let's get together and solve the other remaining questions.
Answered by
robbo
Who's got the capacitor brush problem and the LC circuit???
Answered by
Phy
Q3, 3rd and 4th
I=V/R
I=V/R
Answered by
TA
Phy I think you are not quite right.
Any right answer for the transient
Any right answer for the transient
Answered by
Anonymous
ans for Q3 2
V/R(1-e^-(R*t/L))
5
V/R*e^-(t/(R*C))
V/R(1-e^-(R*t/L))
5
V/R*e^-(t/(R*C))
Answered by
me
Q 7 please
Answered by
aristotle
Q7
look at the maxima peak and the location of the minimum. count off the peaks.
( ist maxima and minima.)
Anyone for the capacitor and sphere problem? Thnks.
look at the maxima peak and the location of the minimum. count off the peaks.
( ist maxima and minima.)
Anyone for the capacitor and sphere problem? Thnks.
Answered by
P
Anonymous what is t in Q3 b
Answered by
LittleJohn
Someone cant help me to resolve a) and c)for the original post Thanks
Answered by
lrc
Answer for a) & c) Q10
Answered by
WTH
answer for c is 3.91
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