Measuring the specific charge of the electron using electric and magnetic fields

Turning Points in Physics | AQA A-Level Physics

Key Definition

Specific charge: The charge-to-mass ratio of a particle. For the electron, $e/m_e = 1.76 \times 10^{11}$ C kg$^{-1}$.

Method 1: Magnetic field only

Electrons are accelerated through a known p.d. $V_a$ and then enter a uniform magnetic field perpendicular to their velocity.
The magnetic force provides the centripetal force for circular motion: $Bev = \dfrac{m_e v^2}{r}$
From the accelerating p.d.: $eV_a = \tfrac{1}{2}m_e v^2$, giving $v = \sqrt{2eV_a / m_e}$
Substituting and rearranging:

$$\frac{e}{m_e} = \frac{2V_a}{B^2 r^2}$$

By measuring $V_a$, $B$, and the radius $r$ of the circular path, the specific charge can be calculated.

Method 2: Balancing electric and magnetic fields (velocity selector)

An electric field $E$ and a magnetic field $B$ are applied perpendicular to each other and to the electron beam.
The electric force $eE$ acts in one direction; the magnetic force $Bev$ acts in the opposite direction.
When the beam is undeflected, the forces balance: $eE = Bev$, so $v = E/B = V/(Bd)$, where $d$ is the plate separation.
The electric field is then switched off. The electrons follow a circular path in the magnetic field alone, with measurable radius $r$.
Combining: $\dfrac{e}{m_e} = \dfrac{v}{Br} = \dfrac{V}{B^2 r d}$

Method 3: Electric field only (parabolic deflection)

If only an electric field is applied between parallel plates, the electron follows a parabolic trajectory (like projectile motion, with horizontal velocity and vertical acceleration).
The vertical acceleration is $a = eE/m_e = eV/(m_e d)$
Using the known plate length, speed, and measured vertical deflection, the specific charge can be calculated.

Common Mistake

In the crossed-fields method, the velocity selector step gives you $v$, but you still need the magnetic-field-only step to get $r$. You cannot find specific charge from the balanced condition alone because both $e$ and $m_e$ cancel out in $v = E/B$.