Question:
Two electrons, \(e_1\) and \(e_2\), of mass \(m\) and charge \(q\) are injected in the perpendicular direction of the magnetic field \(B\) such that the kinetic energy of \(e_1\) is double than that of \(e_2\). The relation of their frequencies of rotation, \(f_1\) and \(f_2\), is
Two electrons, \(e_1\) and \(e_2\), of mass \(m\) and charge \(q\) are injected in the perpendicular direction of the magnetic field \(B\) such that the kinetic energy of \(e_1\) is double than that of \(e_2\). The relation of their frequencies of rotation, \(f_1\) and \(f_2\), is
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Cyclotron frequency \(f=\frac{qB}{2\pi m}\) is independent of speed and kinetic energy. Kinetic energy affects only the radius of the circular path.
Updated On: Jun 15, 2026
- \(f_1=f_2\)
- \(f_1=2f_2\)
- \(2f_1=f_2\)
- \(4f_1=f_2\)
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The Correct Option is A
Solution and Explanation
Step 1: Recall the frequency of rotation of a charged particle in a magnetic field.
When a charged particle enters a uniform magnetic field perpendicular to the field direction, it moves in a circular path.
The frequency of rotation is called cyclotron frequency and is given by
\[ f=\frac{qB}{2\pi m} \]
Step 2: Observe dependence of frequency.
From the formula,
\[ f=\frac{qB}{2\pi m} \]
we can see that the frequency depends only on charge \(q\), magnetic field \(B\), and mass \(m\).
It does not depend on velocity or kinetic energy of the particle.
Step 3: Compare the two electrons.
Both \(e_1\) and \(e_2\) are electrons, so they have the same mass \(m\) and same charge \(q\).
They are moving in the same magnetic field \(B\).
Although the kinetic energy of \(e_1\) is double that of \(e_2\), this changes only the radius of circular motion, not the frequency.
Therefore,
\[ f_1=f_2 \]
Step 4: Final conclusion.
Hence, the relation between their frequencies is
\[ \boxed{f_1=f_2} \]
When a charged particle enters a uniform magnetic field perpendicular to the field direction, it moves in a circular path.
The frequency of rotation is called cyclotron frequency and is given by
\[ f=\frac{qB}{2\pi m} \]
Step 2: Observe dependence of frequency.
From the formula,
\[ f=\frac{qB}{2\pi m} \]
we can see that the frequency depends only on charge \(q\), magnetic field \(B\), and mass \(m\).
It does not depend on velocity or kinetic energy of the particle.
Step 3: Compare the two electrons.
Both \(e_1\) and \(e_2\) are electrons, so they have the same mass \(m\) and same charge \(q\).
They are moving in the same magnetic field \(B\).
Although the kinetic energy of \(e_1\) is double that of \(e_2\), this changes only the radius of circular motion, not the frequency.
Therefore,
\[ f_1=f_2 \]
Step 4: Final conclusion.
Hence, the relation between their frequencies is
\[ \boxed{f_1=f_2} \]
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