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
Learn more about magnetic force :
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