Malus's Law
I=Io•(cosφ)^2,
cos φ =sqrt(I/Io),
φ =19.19o
A horizontal beam of alpha particles are injected with a speed of 1.3x10^5 m/s into a region with a vertical magnetic field of magnitude 0.155T.
A- How long does it take for an alpha particle to move half way through a complete circle?
B- If the speed of the alpha particle is doubled, does the time found in part A increase, decrease, or stay the same? Explain.
C- Repeat for part A for alpha particle with a speed of 2.6x10^5 m/s.
D- What does this infer with regard to the frequency of EMR radiated by identical charges (at any speed) entering a perpendicular magnetic field?
4 answers
I'm sorry. I've mistaken. This is solution of another problem
A. Lorentz force
F=q•v•B•sinα,
sin α =1.
mv^2/R = q•v•B,
R= mv/qB.
The period is
T =2•π•R/v = 2•π•m/q•B.
For α-particle
T = 2•π•4m(p)/2•e•B.
T/2 = 2•π• m(p)/ e•B,
where m(p) = 1.66•10^-27 – the mass of proton
e = 1.6•10^-19 C is the charge of proton.
B. T doesn't depend on velocity.
F=q•v•B•sinα,
sin α =1.
mv^2/R = q•v•B,
R= mv/qB.
The period is
T =2•π•R/v = 2•π•m/q•B.
For α-particle
T = 2•π•4m(p)/2•e•B.
T/2 = 2•π• m(p)/ e•B,
where m(p) = 1.66•10^-27 – the mass of proton
e = 1.6•10^-19 C is the charge of proton.
B. T doesn't depend on velocity.
F=q•v•B•siná,
sin á =1.
mv^2/R = q•v•B,
R= mv/qB.
The period is
T =2•ð•R/v = 2•ð•m/q•B.
For á-particle
T = 2•ð•4m(p)/2•e•B.
T/2 = 2•ð• m(p)/ e•B
sin á =1.
mv^2/R = q•v•B,
R= mv/qB.
The period is
T =2•ð•R/v = 2•ð•m/q•B.
For á-particle
T = 2•ð•4m(p)/2•e•B.
T/2 = 2•ð• m(p)/ e•B