Respuesta :
Answer:
5176.6467 m/s
[tex]5.4\times 10^{-5}\ m[/tex]
231.46352 m
5.40312 m
Explanation:
[tex]m_p[/tex] = Mass of proton = [tex]1.67\times 10^{-27}\ kg[/tex]
[tex]m_e[/tex] = Mass of electron = [tex]9.1\times 10^{-31}\ kg[/tex]
[tex]q_e=q_p[/tex] = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
[tex]v_e[/tex] = Velocity of electron = [tex]9.5\times 10^6\ m/s[/tex]
[tex]v_p[/tex] = Velocity of proton
B = Magnetic field = [tex]1\times 10^{-5}\ T[/tex]
Radius is given by
[tex]r=\frac{mv}{qB}[/tex]
From this relation we get
[tex]\frac{m_pv_p}{q_pB}=\frac{m_ev_e}{q_eB}\\\Rightarrow v_p=\frac{m_ev_e}{m_p}\\\Rightarrow v_p=\frac{9.1\times 10^{-31}\times 9.5\times 10^6}{1.67\times 10^{-27}}\\\Rightarrow v_p=5176.6467\ m/s[/tex]
The speed of proton is 5176.6467 m/s
[tex]r=\frac{1.67\times 10^{-27}\times 9.5\times 10^6}{1.6\times 10^{-19}\times 1\times 10^{-5}}\\\Rightarrow r=9915.625\ m[/tex]
The radius of the path of the electron would be 9915.625 m
For same kinetic energy
[tex]\frac{1}{2}m_ev_e^2=\frac{1}{2}m_pv_p^2\\\Rightarrow v_p=\sqrt{\frac{m_ev_e^2}{m_p}}\\\Rightarrow v_p=\sqrt{\frac{9.1\times 10^{-31}\times (9.5\times 10^6)^2}{1.67\times 10^{-27}}}\\\Rightarrow v_p=221761.45677\ m/s[/tex]
[tex]r=\frac{1.67\times 10^{-27}\times 221761.45677}{1.6\times 10^{-19}\times 1\times 10^{-5}}\\\Rightarrow r=231.46352\ m[/tex]
The radius of the path would be 231.46352 m
For same momentum
[tex]m_ev_e=m_pv_p\\\Rightarrow v_p=\frac{m_ev_e}{m_p}\\\Rightarrow v_p=\frac{9.1\times 10^{-31}\times 9.5\times 10^6}{1.67\times 10^{-27}}\\\Rightarrow v_p=5176.6467\ m/s[/tex]
[tex]r=\frac{1.67\times 10^{-27}\times 5176.6467}{1.6\times 10^{-19}\times 10^{-5}}\\\Rightarrow r=5.40312\ m[/tex]
Radius of the path would be 5.40312 m
(a) The speed of the proton is 5176.6 m/s
(b) The radius of the path be in meters if the proton had the same speed as the electron is 5.4 m.
(c) The radius in meters if the proton had the same kinetic energy as the electron is 231.46 m.
(d) The radius in meters if the proton had the same momentum as the electron is 5.4 m
Movement of a Charged Particle in a Magnetic Field:
(a) The radius of the path of a charge moving in a circular path is obtained using the relation;
[tex]r=\frac{mv}{qB}[/tex]
The radius of the path followed by an electron, we get;
[tex]r_e = \frac{m_e v_e}{eB}= \frac{(9.1\times 10^{-31}\,kg)\times(9.5\times 10^6 \,m/s)}{(1.6\times 10^{-19}\,C)\times (10^{-5}\,T)} = 5.403\,m[/tex]
But, it is given that the proton moves in a circular path of the same radius as the electron here.
So, [tex]r_e = r_p = 5.4\,m[/tex]
Now, rewriting the relation for the radius of the path for proton, we get;
[tex]r_p = \frac{m_p v_p}{eB}[/tex]
[tex]\implies v_p = \frac{r_p \,eB}{m_p}=\frac{5.4\,m \times 1.6 \times 10^{-19}\.C \times 10^{-5}\,T}{1.67\times 10^{-27}\,kg}= 5176.6\,m/s[/tex]
(b) If [tex]v_p = v_e = 9.5 \times 10^6\, m/s[/tex], then;
[tex]r_p = \frac{(1.67\times 10^{-27}\,kg)\times )(9.5\times 10^6 \,m/s)}{(1.6\times 10^{-19}C)\times (10^{-5}\,T)} =9915.625\,m/s[/tex]
(c) If [tex](KE)_{electron}=(KE)_{proton}[/tex];
[tex]\frac{1}{2} m_e(v_e)^2 = \frac{1}{2} m_p(v_p)^2[/tex]
[tex]v_p=\sqrt{\frac{m_e (v_e)^2}{m_p} }[/tex]
Substituting the known values, we get;
[tex]v_p=\sqrt{\frac{(9.1\times 10^{-31}\,kg) (9.5\times 10^6\,m/s)^2}{1.67\times10^{-27}\,kg} }=221761.45\,m/s[/tex]
Now;
[tex]r_p = \frac{m_p v_p}{eB}= \frac{(1.67\times 10^{-27}\,kg)\times(221761 \,m/s)}{(1.6\times 10^{-19}\,C)\times (10^{-5}\,T)} = 231.46\,m[/tex]
(d) GIven that the proton has the same momentum as the electron.
[tex]m_p v_p = m_e v_e[/tex]
[tex]\implies v_p = \frac{m_e v_e}{m_p} = \frac{(9.1\times 10^{-31}\,kg)\times(9.5\times 10^6\,m/s)}{1.67\times 10^{-27}kg} =5176.64\,m/s[/tex]
Therefore;
[tex]r_p = \frac{m_p v_p}{eB}= \frac{(1.67\times 10^{-27}\,kg)\times(5176.64 \,m/s)}{(1.6\times 10^{-19}\,C)\times (10^{-5}\,T)} = 5.4 \,m[/tex]
Learn more about the movement of a charged particle in a magnetic field here:
https://brainly.com/question/25798298
