A calcium-40 ion has a positive charge that is double the charge of a proton, and a mass of 6.64 ✕ 10−26 kg. At a particular ins
tant, it is moving with a speed of 5.00 ✕ 106 m/s through a magnetic field. At this instant, its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location. The magnitude of the field is 0.230 T.
(a)What is the magnitude of the magnetic force (in N) on the ion?
N
B)What is the magnitude of the ion's acceleration (in m/s2) at this instant?
m/s2
(a)What is the magnitude of the magnetic force (in N) on the ion?
N
B)What is the magnitude of the ion's acceleration (in m/s2) at this instant?
m/s2
1 answer:
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Answer:
F = 6666.7 N
Explanation:
Given that,
Mass of a chip, m = 0.1 mg
Initial speed, u = 0
Final speed,
Time of collision,
We know that,
Force, F = ma
Put all the values,
So, the required force is 6666.7 N.