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?

Answer :

Answer:

The value is [tex]B = 7.7106 *10^{-12 } \ T[/tex]

Explanation:

From the question we are told that

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

The  radius  is  [tex]R = 5.80 *10^{10} \ m[/tex]

Generally the magnetic field is mathematically represented as

       [tex]B = \frac{m * v }{q * R }[/tex]

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

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

                   [tex]v = \sqrt{ \frac{E }{ 0.5 * m } }[/tex]

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

                   [tex]v = \sqrt{ \frac{1.6 *10^{-12} }{ 0.5 * 1.6*10^{-27} } }[/tex]

                 [tex]v = 4.47214 *10^{7} \ m/s[/tex]

So  

     [tex]B = \frac{1.60 *10^{-27} * 4.47214 *10^{7}}{ 1.60 *10^{-19} * 5.80*10^{10}}[/tex]

      [tex]B = 7.7106 *10^{-12 } \ T[/tex]

The magnetic field in the region has a magnitude of 7.7×[tex]10^{-12}[/tex] T.

Let's consider the magnetic field in the region be B. Now the proton in the given orbit is in equilibrium under the influence of centripetal force ([tex]F_{c}[/tex]) and magnetic force([tex]F_{b}[/tex]).

[tex]F_{c}[/tex] = [tex]F_{b}[/tex]

m[tex]v^{2}[/tex]/R = qvB

⇒ B = mv/qR

where, m is the mass of proton =  1.6 × [tex]10^{-27}[/tex] kg

            q is the charge of proton = 1.6 × [tex]10^{-19}[/tex] C

            R is the radius of the orbit = 5.8 × [tex]10^{10}[/tex] m

now we can calculate the velocity of the proton, v, from the energy of the proton given in the question that is E = 10 MeV.

1 MeV = 1.6 × [tex]10^{-19}[/tex] J

           v = [tex]\sqrt{\frac{2E}{m} }[/tex]

               =[tex]\sqrt{\frac{2*10*10^{6}* 1.6*10^{-19} }{1.6*10^{-27} } }[/tex]

            v = 4.47 × [tex]10^{7}[/tex] m/s

now,

        B = mv/qR

           = [tex]\frac{1.6*10^{-27}*4.47*10^{7} }{1.6*10^{-19}*5.8*10^{10} }[/tex]

       B = 7.7 ×[tex]10^{-12}[/tex] T

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