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Transcript
Please review my solution to the problem and explain in detail what I may be doing wrong and what concepts I
may not be applying correctly.
The problem states:
An alpha particle has a charge of +3.2 x 10 -19 C and a mass of 6.6 x 10-27 kg. The alpha particle travels at a
velocity v of magnitude 550 m/s through a uniform magnetic field B of magnitude 0.045T. The angle between v
and B is 52º.
a) What is the magnitude of the force FB acting on the particle due to the field?
b) What is the acceleration of the particle due to FB?
c) Does the speed of the particle increase, decrease, or remain the same?
We know:
V = 550 m/s
B = 0.045T
Q = +3.2 x 10-19 C
Mass = 6.6 x 10-27 kg
Φ = 52º
Solution:
a)
B = 0.045T = 0.045 N/Cm/s
FB = qvB
FB = (+3.2 x 10-19 C)( 550 m/s)( 0.045 N/Cm/s)
FB = 7.92 x 10-18 N
b)
FB = │q│vBsinΦ
sinΦ = sin(52º) = .7880
v = 7.92 x 10-18 N / [(3.2 x 10-19 C)( 0.045 N/Cm/s)( .7880)
v = 6.9797 x 102 m/s
c)
The speed stays the same because FB cannot change the particles speed, just direction.