
If a proton and an $\alpha $ particle enter in a uniform magnetic field with the same velocity, what will be the period of rotation of the proton?
A. One-fourth of that of the $\alpha $ particle.
B. Half of that of the $\alpha $ particle.
C. One-third of that of the $\alpha $particle.
D. Same as that of the $\alpha $ particle.
Answer
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Hint: In the case of a problem based on the magnetic field, we know that period of rotation inversely varies with the uniform magnetic field. Also, there is a relationship between mass and charge of both proton and $\alpha$ particle which helps in establishing a relation between the period of rotation of proton and $\alpha $particle.
Formula used:
Time period of rotation in a uniform magnetic field is given as:
$T = \dfrac{{2\pi m}}{{Bq}}$
Here, T = time-period of rotation
B = magnetic field intensity
Complete step by step solution:
Let ${m_p} = $ mass of proton, ${q_p} = $ charge on proton, ${m_\alpha } = $ mass of $\alpha $ particle, and ${q_\alpha } = $ charge on $\alpha $particle. Now, we know that the time period of rotation in a uniform magnetic field is given as:
$T = \dfrac{{2\pi m}}{{Bq}}$
Applying this formula for proton, we get
${T_p} = \dfrac{{2\pi {m_p}}}{{B{q_p}}}$ … (1)
For $\alpha $particle,
${T_\alpha } = \dfrac{{2\pi {m_\alpha }}}{{B{q_\alpha }}}$ … (2)
B is same for both cases as the magnetic field is uniform and both proton and $\alpha $ particle enter in the same magnetic field.
Also, we know that
${m_\alpha } = 4{m_p}$ and ${q_\alpha } = 2{q_p}$
From eq. (2), we get
\[{T_\alpha } = \dfrac{{2\pi \left( {4{m_p}} \right)}}{{B\left( {2{q_p}} \right)}} = 2.\dfrac{{2\pi {m_p}}}{{B{q_p}}} \\ \] … (3)
From eq. (3) and (1), we get
\[ \Rightarrow {T_\alpha } = 2{T_p}\]
\[ \therefore {T_p} = \dfrac{1}{2}{T_\alpha }\]
Thus, the time period of rotation of the proton is Half of that of the $\alpha $particle.
Hence, the correct option is B.
Note: Since this is a problem related to uniform magnetic field hence, given conditions are to be analyzed very carefully and quantities that are required to calculate the time period must be identified on a prior basis as it gives a better understanding of the problem and helps to further solve the question.
Formula used:
Time period of rotation in a uniform magnetic field is given as:
$T = \dfrac{{2\pi m}}{{Bq}}$
Here, T = time-period of rotation
B = magnetic field intensity
Complete step by step solution:
Let ${m_p} = $ mass of proton, ${q_p} = $ charge on proton, ${m_\alpha } = $ mass of $\alpha $ particle, and ${q_\alpha } = $ charge on $\alpha $particle. Now, we know that the time period of rotation in a uniform magnetic field is given as:
$T = \dfrac{{2\pi m}}{{Bq}}$
Applying this formula for proton, we get
${T_p} = \dfrac{{2\pi {m_p}}}{{B{q_p}}}$ … (1)
For $\alpha $particle,
${T_\alpha } = \dfrac{{2\pi {m_\alpha }}}{{B{q_\alpha }}}$ … (2)
B is same for both cases as the magnetic field is uniform and both proton and $\alpha $ particle enter in the same magnetic field.
Also, we know that
${m_\alpha } = 4{m_p}$ and ${q_\alpha } = 2{q_p}$
From eq. (2), we get
\[{T_\alpha } = \dfrac{{2\pi \left( {4{m_p}} \right)}}{{B\left( {2{q_p}} \right)}} = 2.\dfrac{{2\pi {m_p}}}{{B{q_p}}} \\ \] … (3)
From eq. (3) and (1), we get
\[ \Rightarrow {T_\alpha } = 2{T_p}\]
\[ \therefore {T_p} = \dfrac{1}{2}{T_\alpha }\]
Thus, the time period of rotation of the proton is Half of that of the $\alpha $particle.
Hence, the correct option is B.
Note: Since this is a problem related to uniform magnetic field hence, given conditions are to be analyzed very carefully and quantities that are required to calculate the time period must be identified on a prior basis as it gives a better understanding of the problem and helps to further solve the question.
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