Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.
Give your answer in centimetres.
1 answer:
Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r =
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r = = 0.0328 m, or 3.28 cm.
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position as a function of time is y = 0.05 × cos(9.9)t
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mass = 5 kg
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