
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
216k+ views
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.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

Alpha, Beta, and Gamma Decay Explained

Alpha Particle Scattering and Rutherford Model Explained

Angular Momentum of a Rotating Body: Definition & Formula

Apparent Frequency Explained: Formula, Uses & Examples

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

How to Convert a Galvanometer into an Ammeter or Voltmeter

Atomic Structure: Definition, Models, and Examples

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

