Physics, 23.08.2020 01:01 hjeffrey168

A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the Sun (5.80 × 1010m). What is the magnetic field in that region of space?

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Answer from: coolusername1314

The value is $B = 7.7106 *10^{-12 } \ T$

Explanation:

From the question we are told that

The energy is $E = 10.0 MeV = 10.0 *10^{6} \ eV = 10.0 *10^{6} * 1.60*10^{-19} = 1.6 *10^{-12} \ J$

The radius is $R = 5.80 *10^{10} \ m$

Generally the magnetic field is mathematically represented as

$B = \frac{m * v }{q * R }$

Where m is the mass of proton with value $m = 1.6*10^{-27} \ kg$

v is the velocity of the proton which is mathematically deduced from the formula for kinetic energy as

$v = \sqrt{ \frac{E }{ 0.5 * m } }$

Here E is also equivalent to kinetic energy of the proton so

$v = \sqrt{ \frac{1.6 *10^{-12} }{ 0.5 * 1.6*10^{-27} } }$

$v = 4.47214 *10^{7} \ m/s$

$B = \frac{1.60 *10^{-27} * 4.47214 *10^{7}}{ 1.60 *10^{-19} * 5.80*10^{10}}$

$B = 7.7106 *10^{-12 } \ T$

Answer from: Quest

Centrapedial force is the answer

Answer from: Quest

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