
An electron and a proton with equal momentum enter perpendicularly into a uniform magnetic field, then
A. The path of proton shall be more curved than that of electron
B. The path of proton shall be less curved than that of electron
C. Both are equally curved
D. Path of both will be straight line
Answer
164.1k+ views
Hint: When a charged particle enters into a region of perpendicular magnetic field of uniform magnetic field strength then it experiences a magnetic force in the direction perpendicular to the initial velocity. So the resulting path of the motion is circular in nature.
Formula used:
\[r = \dfrac{{mv}}{{Bq}}\], here r is the radius of the circular path when a charged particle of mass m and charge q enters into a region of magnetic field strength B with speed v.
\[p = mv\], here p is the magnitude of the momentum of the body of mass m moving with speed v.
Complete answer:
Let the mass of the electron is \[{m_e}\]and the mass of the proton is \[{m_p}\]
The charge on the electron and the proton is same in magnitude \[{q_e} = {q_p} = e\]
It is given that the electron and proton is moving with equal momentum,
\[{m_e}{v_e} = p\]
\[{m_p}{v_p} = p\]
On dividing both the expressions of the momentum, we get
\[\dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{{m_p}}}{{{m_e}}}\]
Using the radius formula,
The radius of the circular path of the electron is,
\[{r_e} = \dfrac{{{m_e}{v_e}}}{{Be}}\]
\[{r_e} = \dfrac{p}{{Be}}\]
The radius of the circular path of the proton is,
\[{r_p} = \dfrac{{{m_p}{v_p}}}{{Be}}\]
\[{r_p} = \dfrac{p}{{Be}}\]
On dividing both the expressions for the radius of the circular path,
\[\dfrac{{{r_e}}}{{{r_p}}} = \dfrac{{\left( {\dfrac{p}{{Be}}} \right)}}{{\left( {\dfrac{p}{{Be}}} \right)}}\]
\[{r_e} = {r_p}\]
So, the radius of the circular path for the electron is equal to the radius of the circular path for the proton.
Therefore, the correct option is (C).
Note:As the nature of the charge on proton and electron is opposite, so the direction of the deflection will be opposite to each other though the radius of the circular path will be equal.
Formula used:
\[r = \dfrac{{mv}}{{Bq}}\], here r is the radius of the circular path when a charged particle of mass m and charge q enters into a region of magnetic field strength B with speed v.
\[p = mv\], here p is the magnitude of the momentum of the body of mass m moving with speed v.
Complete answer:
Let the mass of the electron is \[{m_e}\]and the mass of the proton is \[{m_p}\]
The charge on the electron and the proton is same in magnitude \[{q_e} = {q_p} = e\]
It is given that the electron and proton is moving with equal momentum,
\[{m_e}{v_e} = p\]
\[{m_p}{v_p} = p\]
On dividing both the expressions of the momentum, we get
\[\dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{{m_p}}}{{{m_e}}}\]
Using the radius formula,
The radius of the circular path of the electron is,
\[{r_e} = \dfrac{{{m_e}{v_e}}}{{Be}}\]
\[{r_e} = \dfrac{p}{{Be}}\]
The radius of the circular path of the proton is,
\[{r_p} = \dfrac{{{m_p}{v_p}}}{{Be}}\]
\[{r_p} = \dfrac{p}{{Be}}\]
On dividing both the expressions for the radius of the circular path,
\[\dfrac{{{r_e}}}{{{r_p}}} = \dfrac{{\left( {\dfrac{p}{{Be}}} \right)}}{{\left( {\dfrac{p}{{Be}}} \right)}}\]
\[{r_e} = {r_p}\]
So, the radius of the circular path for the electron is equal to the radius of the circular path for the proton.
Therefore, the correct option is (C).
Note:As the nature of the charge on proton and electron is opposite, so the direction of the deflection will be opposite to each other though the radius of the circular path will be equal.
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