Physics, 10.08.2019 03:10 awkwardness92

Aproton, which moves perpendicular to a magnetic field of 1.35 t in a circular path of radius 0.072 m, has what speed? (qp = 1.6 · 10 19 c and mp = 1.67 · 10 27 kg)

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Answer from: kuuleilani

Speed of the proton is $v=9.31\times 10^6\ m/s$

Explanation:

It is given that,

Magnetic field, B = 1.35 T

Radius of circular path, r = 0.072 m

The centripetal force is balanced by the magnetic force such that,

$\dfrac{mv^2}{r}=qv$

$v=\dfrac{qrB}{m}$

q and m is the charge and mass of a proton respectively

$v=\dfrac{1.6\times 10^{-19}\times 0.072\times 1.35}{1.67\times 10^{-27}}$

$v=9.31\times 10^6\ m/s$

So, the speed if the proton is $9.31\times 10^6\ m/s$ . Hence, this is the required solution.

Answer from: Quest

i dont think you are wrong for wanting to get even with someone who did you wrong.

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Answer from: Quest

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Aproton, which moves perpendicular to a magnetic field of 1.35 t in a circular path of radius 0.072...

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