Q.In a chamber, a uniform magnetic field of $6.5\ \text{G}$ ($1\ \text{G} = 10^{-4}\ \text{T}$) is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6}\ \text{m s}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. ($e = 1.5 \times 10^{-19}\ \text{C}$, $m_e = 9.1 \times 10^{-31}\ \text{kg}$).

Concept: charged particle in a $\perp$ magnetic field — circular motion.

Why a circle: the force $\vec F=q(\vec v\times\vec B)$ is always perpendicular to $\vec v$, so it does no work — the speed is constant. With $\vec v\perp\vec B$ this constant-magnitude force always points to one centre, i.e. it is centripetal, giving uniform circular motion.

Radius: from $evB=\dfrac{m_e v^2}{r}$, $\ r=\dfrac{m_e v}{eB}$. With $B=6.5\ \text{G}=6.5\times10^{-4}\ \text{T}$ and the stated $e=1.5\times10^{-19}\ \text{C}$: …

Why the path is a circle

The electron feels only the magnetic force $\vec F=q(\vec v\times\vec B)$, which is always perpendicular to the velocity. A force perpendicular to $\vec v$ does no work, so the kinetic energy — and hence the speed — never changes. Since the electron enters normal to $\vec B$, this force has constant magnitude $evB$ and always points toward one fixed centre: exactly the condition for uniform circular motion.

Determining the radius

The magnetic force provides the centripetal force:

Convert the field: $B=6.5\ \text{G}=6.5\times10^{-4}\ \text{T}$. Substituting the stated data ($m_e=9.1\times10^{-31}\ \text{kg}$, $v=4.8\times10^{6}\ \text{m s}^{-1}$, $e=1.5\times10^{-19}\ \text{C}$):

Numerator: $9.1\times4.8=43.68$, so $43.68\times10^{-25}$. Denominator: $1.5\times6.5=9.75$, so $9.75\times10^{-23}$. Hence …

Method: Lorentz Force & Centripetal Force Equivalence

Why the path is a circle

When a charged particle moves perpendicular to a uniform magnetic field:

• The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ acts perpendicular to both velocity and field
• Since $\vec{v} \perp \vec{B}$, the force magnitude is $F_m = |q|vB$
• This force is always perpendicular to velocity → it changes only the direction, not the speed
• A constant perpendicular force causes uniform circular motion

Steps to find the radius

Step 1: Equate forces

The magnetic force provides the centripetal force:

Step 2: Solve for radius

Cancel $v$ from both sides:

Step 3: Convert units

$B = 6.5\ \text{G} = 6.5 \times 10^{-4}\ \text{T}$

Step 4: Substitute values

$m_e = 9.1 \times 10^{-31}\ \text{kg}$, $v = 4.8 \times 10^{6}\ \text{m/s}$, $e = 1.5 \times 10^{-19}\ \text{C}$

Step 5: Calculate

Numerator: $9.1 \times 4.8 \times 10^{-25} = 43.68 \times 10^{-25}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting to Convert Gauss to Tesla

The error: Students plug $B = 6.5\ \text{G}$ directly into formulas, forgetting the conversion factor.

Why it happens: The problem gives $B$ in Gauss but all standard formulas use Tesla. The conversion hint ($1\ \text{G} = 10^{-4}\ \text{T}$) is easy to miss under time pressure.

How to avoid: Always write the conversion step explicitly:

Pro tip: Circle or underline the conversion factor in the question before starting calculations.

Mistake 2: Using Wrong Charge Value

The error: Using $e = 1.6 \times 10^{-19}\ \text{C}$ (the standard value) instead of the given $e = 1.5 \times 10^{-19}\ \text{C}$.

Why it happens: Students memorize the standard electron charge and automatically substitute it without checking the problem's data.

How to avoid: Always use the values provided in the question, even if they differ from standard textbook values. The problem deliberately gives $1.5 \times 10^{-19}\ \text{C}$ — use it.

Mistake 3: Confusing the Reason for Circular Motion

The error: Saying "the electron moves in a circle because the magnetic force is perpendicular to velocity" — but not explaining why this produces a circle.

Why it happens: Students memorize the result without understanding the mechanism.

How to avoid: Explain step-by-step:

• Magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ acts perpendicular to both $\vec{v}$ and $\vec{B}$
• Since $\vec{v} \perp \vec{B}$, the force magnitude is $F = qvB$
• This perpendicular force provides centripetal acceleration $a_c = v^2/r$
• The force changes only the direction of velocity, not its magnitude
• Result: uniform circular motion

Mistake 4: Sign Errors in Force Direction

The error: Forgetting that the electron has negative charge, so the force direction is opposite to that for a positive charge.

Why it happens: Students apply the right-hand rule for positive charges without flipping the direction for electrons.

How to avoid: Remember: For electrons, use left-hand rule or apply the right-hand rule and then reverse the direction. The magnitude calculation is unaffected, but conceptual questions about direction will be wrong.

Mistake 5: Formula Confusion — Radius Expression

The error: Writing $r = \frac{mv}{qB}$ incorrectly as $r = \frac{mv^2}{qB}$ or $r = \frac{qB}{mv}$.

Why it happens: Mixing up centripetal force ($mv^2/r$) with magnetic force ($qvB$).

How to avoid: Derive it quickly:

• Centripetal force = Magnetic force
• $\frac{mv^2}{r} = qvB$
• Cancel one $v$: $\frac{mv}{r} = qB$ …