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natali 33 [55]
3 years ago
6

A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of

0.00687 T 0.00687 T and the proton's velocity makes an angle of 101 ∘ 101∘ with the field, what is the magnitude of the magnetic force acting on the proton?
Physics
1 answer:
iogann1982 [59]3 years ago
4 0

Answer:

8.64283\times 10^{-14}\ N

Explanation:

q = Charge of proton = 1.6\times 10^{-19}\ C

v = Velocity of proton = 0.267\times c

c = Speed of light = 3\times 10^8\ m/s

B = Magnetic field = 0.00687 T

\theta = Angle = 101^{\circ}

Magnetic force is given by

F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N

The magnetic force acting on the proton is 8.64283\times 10^{-14}\ N

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When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:

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