Q.In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

The key idea is that a charged particle moving perpendicular to a uniform magnetic field experiences a centripetal force, giving circular motion with a frequency independent of speed.

Reasoning:

1. For an electron of mass $m$ and charge magnitude $e$ moving with speed $v$ perpendicular to a field $B$, the magnetic force $evB$ supplies the centripetal force:

2. The period of one revolution is

3. The frequency is

Notice $v$ has cancelled out completely -- $f$ depends only on $e$, $B$, and $m$, never on the electron's speed. …

Why This Works: The Concept

When a charged particle like an electron moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force, bending the path into a circle. The magnetic force depends on speed ($evB$), but so does the centripetal requirement ($mv^2/r$) -- these two speed-dependences cancel when solving for the period, leaving a frequency that depends only on the charge-to-mass ratio and the field strength. This is exactly why cyclotrons work: particles of different speeds still complete one revolution in the same time.

Step-by-Step Solution

1. Set up the force balance. For an electron of mass $m$ and charge magnitude $e$ moving with speed $v$ perpendicular to a uniform field $B$, the magnetic force supplies the centripetal force:

2. Solve for the radius.

A faster electron traces a larger circle.

3. Find the period $T$.

The speed $v$ cancels out completely.

4. Obtain the frequency.

5. Compute the numeric value, continuing the earlier exercise's data. That exercise gives $B = 6.5\ \text{G} = 6.5\times10^{-4}\ \text{T}$, with $e = 1.6\times10^{-19}\ \text{C}$ and $m_e = 9.1\times10^{-31}\ \text{kg}$:

So the electron completes about $1.82\times10^{7}$ revolutions every second (roughly $18\ \text{MHz}$).

6. Does the answer depend on speed? …

Method: Centripetal Force Equals Magnetic Lorentz Force

This is the standard method for finding the cyclotron frequency (or gyrofrequency) of a charged particle moving perpendicular to a uniform magnetic field.

Steps

1. Identify the force providing centripetal acceleration For an electron moving in a circle of radius $r$ with speed $v$, the centripetal force required is:

where $m$ is the electron's mass.

2. Identify the magnetic force on the moving charge For a charge $q$ moving with velocity $\vec{v}$ perpendicular to a uniform magnetic field $\vec{B}$, the magnetic Lorentz force is:

(The direction is given by the right-hand rule, but magnitude is what matters here.)

3. Set the forces equal (since the magnetic force provides the centripetal force):

4. Solve for the radius (optional, but helps find frequency):

5. Relate speed to angular frequency For circular motion, $v = \omega r$, where $\omega$ is the angular frequency (in rad/s). Substitute into the radius equation:

6. Cancel $r$ (assuming $r \neq 0$) and solve for $\omega$:

7. Convert to frequency of revolution (cycles per second):

Final Answer

For an electron, $|q| = e$ (magnitude of electron charge), so: …

Common Mistakes: Charged Particle in Magnetic Field (Exercise 4.11)

Mistake #1: Forgetting the Formula for Frequency

The error: Students often confuse frequency ($\nu$) with angular frequency ($\omega$) or use the wrong expression.

Correct approach:

For a charged particle moving in a uniform magnetic field:

• Centripetal force is provided by magnetic force:

• Radius of orbit:

• Time period ($T$) = $\frac{2\pi r}{v} = \frac{2\pi m}{q B}$

• Frequency of revolution:

Key insight: The frequency depends only on $q$, $B$, and $m$ — not on speed $v$.

Mistake #2: Thinking Frequency Depends on Speed

The error: Many students assume that a faster electron means more revolutions per second.

Why it's wrong:

• A faster electron has a larger radius ($r \propto v$), so it travels a longer circumference in the same time.
• The increase in path length exactly cancels the increase in speed.
• Result: Time period (and frequency) is independent of speed.

How to avoid: Always derive $T = \frac{2\pi m}{qB}$ and notice $v$ cancels out.

Mistake #3: Using Wrong Units or Constants

The error:

• Using mass of electron in grams instead of kg
• Forgetting $q = 1.6 \times 10^{-19}$ C
• Mixing up $m$ (mass) with $m$ (metres)

How to avoid:

• Write all quantities in SI units before substituting
• Double-check: mass in kg, charge in C, magnetic field in Tesla

Mistake #4: Confusing Frequency with Angular Frequency

The error: Stating $\omega = \frac{qB}{2\pi m}$ instead of $\omega = \frac{qB}{m}$

Correct relationships:

How to avoid: Remember $\omega = 2\pi \nu$, so if you derive $\omega$, divide by $2\pi$ to get $\nu$.

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