An electron is shot into the uniform magnetic field, normal to the direction of field. Then the frequency of revolution of the electron in its circular orbit :
Electron Motion in a Uniform Magnetic Field
When an electron, which is a charged particle, is shot into a uniform magnetic field in a direction normal (perpendicular) to the field lines, it experiences a magnetic force. This magnetic force acts perpendicular to both the velocity of the electron and the magnetic field direction. Because the force is always perpendicular to the velocity, it acts as a centripetal force, causing the electron to move in a circular path.
Forces Involved in Circular Motion
For the electron to move in a stable circular orbit, the magnetic force provides the necessary centripetal force.
- Magnetic Force (\(F_B\)): The magnitude of the magnetic force on a charged particle moving in a magnetic field is given by \(F_B = |q|vB \sin\theta\). Here, \(|q|\) is the magnitude of the charge (for an electron, it's \(e\)), \(v\) is the speed of the electron, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity vector and the magnetic field vector. In this problem, the electron is shot normal to the field, so \(\theta = 90^\circ\), and \(\sin 90^\circ = 1\). Thus, the magnetic force is \(F_B = evB\).
- Centripetal Force (\(F_C\)): For an object moving in a circle of radius \(r\) with speed \(v\), the required centripetal force is given by \(F_C = \frac{mv^2}{r}\), where \(m\) is the mass of the electron.
Deriving the Frequency of Revolution
Equating the magnetic force and the centripetal force for the circular motion:
\(F_B = F_C\)
\(evB = \frac{mv^2}{r}\)
From this equation, we can find the radius of the circular path:
\(r = \frac{mv^2}{evB}\)
\(r = \frac{mv}{eB}\)
This formula shows that the radius of the orbit is directly proportional to the speed of the electron. A faster electron will move in a larger circle.
The frequency of revolution (\(f\)) is the number of circles completed per unit time. It is related to the angular velocity (\(\omega\)) by \(f = \frac{\omega}{2\pi}\). The angular velocity is also related to the linear speed (\(v\)) and radius (\(r\)) by \(\omega = \frac{v}{r}\).
Substituting the expression for \(r\) into the formula for \(\omega\):
\(\omega = \frac{v}{(mv/eB)}\)
\(\omega = \frac{v \cdot eB}{mv}\)
\(\omega = \frac{eB}{m}\)
Now, substituting this into the formula for frequency \(f\):
\(f = \frac{\omega}{2\pi}\)
\(f = \frac{eB}{2\pi m}\)
Analyzing the Frequency Formula
The formula for the frequency of revolution is \(f = \frac{eB}{2\pi m}\). Let's look at the terms in this formula:
- \(e\): This is the magnitude of the charge of the electron, a constant value.
- \(B\): This is the strength of the uniform magnetic field, which is constant in this problem.
- \(m\): This is the mass of the electron, a constant value.
- \(2\pi\): This is a mathematical constant.
As we can see, the frequency \(f\) depends only on the charge of the particle, the magnetic field strength, and the mass of the particle. It does not depend on the speed (\(v\)) of the electron or the radius (\(r\)) of its orbit. This frequency is often called the cyclotron frequency because it is the operating frequency of a cyclotron particle accelerator.
Evaluating the Options
Based on our derivation, let's consider the given options for the frequency of revolution of the electron:
- Option 1: is independent of its speed
Our formula \(f = \frac{eB}{2\pi m}\) clearly shows that the speed \(v\) does not appear in the expression for \(f\). Thus, the frequency is independent of the speed. This option aligns with our findings. - Option 2: decreases with its speed
This contradicts our derived formula. The frequency does not depend on speed at all in this classical treatment. - Option 3: increases with its speed
This also contradicts our derived formula. The frequency does not depend on speed. - Option 4: increase with radius of revolution
The frequency is \(f = \frac{eB}{2\pi m}\), which does not contain \(r\). While the radius \(r = \frac{mv}{eB}\) depends on speed (and thus indirectly related to the energy which affects speed), the frequency itself is independent of both speed and radius.
Therefore, the frequency of revolution of the electron in its circular orbit in a uniform magnetic field, when shot normal to the field, is independent of its speed.
Revision Table: Electron in Magnetic Field
| Concept | Formula/Description | Dependency on Speed/Radius |
|---|---|---|
| Magnetic Force ($F_B$) | \(F_B = evB\) (for \(\theta = 90^\circ\)) | Depends on speed ($v$) |
| Centripetal Force ($F_C$) | \(F_C = \frac{mv^2}{r}\) | Depends on speed ($v$) and radius ($r$) |
| Equating Forces ($F_B = F_C$) | \(evB = \frac{mv^2}{r}\) | Relates $v$ and $r$ |
| Radius of Orbit ($r$) | \(r = \frac{mv}{eB}\) | Directly proportional to speed ($v$). Independent of magnetic field strength ($B$) and mass ($m$). |
| Angular Velocity ($\omega$) | \(\omega = \frac{eB}{m}\) | Independent of speed ($v$) and radius ($r$) |
| Frequency of Revolution ($f$) | \(f = \frac{eB}{2\pi m}\) | Independent of speed ($v$) and radius ($r$) |
Additional Information: Cyclotron Frequency and Related Concepts
The frequency \(f = \frac{qB}{2\pi m}\) derived for a charged particle (like an electron or proton) moving perpendicular to a uniform magnetic field is a very important concept in physics. It is known as the cyclotron frequency or gyrofrequency. This frequency is constant for a given type of particle and magnetic field strength, regardless of the particle's energy or speed (as long as the speed is not relativistic).
Why is the Frequency Constant?
Think about it this way: as the speed of the electron increases, the magnetic force ($F_B = evB$) increases. This stronger force requires a larger centripetal force. For a given speed, a larger centripetal force is achieved by moving in a larger circle. So, the radius of the orbit ($r = \frac{mv}{eB}$) increases proportionally with the speed. Since both the circumference (which is proportional to the radius) and the speed increase by the same factor, the time taken to complete one revolution (\(T = \frac{2\pi r}{v}\)) remains constant:
\(T = \frac{2\pi (mv/eB)}{v} = \frac{2\pi mv}{evB} = \frac{2\pi m}{eB}\)
The frequency, being the reciprocal of the time period ($f = 1/T$), is therefore also constant:
\(f = \frac{eB}{2\pi m}\)
Relativistic Effects
The derivation above uses classical mechanics. At very high speeds, close to the speed of light, relativistic effects become significant. The mass of the particle effectively increases with speed (\(m = \gamma m_0\), where \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}\) and \(m_0\) is the rest mass). In the relativistic case, the frequency formula becomes \(f = \frac{eB}{2\pi \gamma m_0}\). This means that at relativistic speeds, the frequency *does* decrease as the speed increases (because \(\gamma\) increases). However, for typical electron speeds in many introductory physics problems, the classical approximation is valid.
Applications
The principle of constant cyclotron frequency is fundamental to the operation of cyclotrons and other particle accelerators, which use oscillating electric fields synchronized with the particle's revolution frequency to repeatedly accelerate the particles.
Important Questions from Motion in a Magnetic Field
Which of the following is the correct expression for cyclotron frequency?
The radius of curvature of a charged particle's passage in a homogeneous magnetic field is proportional to:
What is the period of revolution T of charge - q1 if a particle of mass m carrying a charge - q1 starts moving around a fixed charge + q2 along a circular path of radius r?
A proton and an alpha particle moving with same kinetic energy enter in the region of uniform magnetic field perpendicular to it. The ratio of radii of their trajectories will be:
- A 2.0 m long metallic rod is rotated with an angular frequency of 100 rad/s about an axis normal to the rod and passing through its one end. A uniform magnetic field of 2.0 T exits parallel to the axis. The emf induced between the two ends of the rod is:
