Question

A beam of non-relativistic charged particles moves without deviation through the region of space A where there are transverse mutually perpendicular electric and magnetic fields with strength E and induction B. When the magnetic field is switched off, the trace of the beam on the screen S shifts by $\Delta x$. Knowing the distance a and b, find the specific charge $q/m$ of the particles.

Answer 1

$\frac{2 \Delta x E}{aB^2 (a+2b)}$

Answer 2

The problem describes two scenarios for the motion of non-relativistic charged particles:

Scenario 1: Velocity Selector (E and B fields present)

A beam of charged particles moves without deviation through a region of space A where there are mutually perpendicular electric field (E) and magnetic field (B). For the particles to move undeflected, the electric force ($F_e = qE$) must balance the magnetic force ($F_m = qvB$).

$qE = qvB$

From this, the velocity of the particles ($v$) is determined:

$v = \frac{E}{B}$

This is the principle of a velocity selector.

Scenario 2: Deflection by Electric Field (B field switched off)

When the magnetic field is switched off, only the electric field E acts on the particles in region A. The electric field is transverse to the initial velocity of the particles. Let the length of region A be 'a'. The time taken for a particle to traverse region A is $t_A = \frac{a}{v}$. During this time, the particle experiences a constant electric force $F_e = qE$ in the transverse direction. This force causes an acceleration $a_y = \frac{F_e}{m} = \frac{qE}{m}$ in the transverse direction.

1. Deflection within region A:

The transverse displacement ($y_A$) of the particle when it exits region A is:

$y_A = \frac{1}{2} a_y t_A^2 = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{a}{v}\right)^2 = \frac{qEa^2}{2mv^2}$

2. Transverse velocity at exit from region A:

The transverse velocity ($v_y$) acquired by the particle when it exits region A is:

$v_y = a_y t_A = \left(\frac{qE}{m}\right) \left(\frac{a}{v}\right) = \frac{qEa}{mv}$

3. Deflection after exiting region A:

After exiting region A, the particles travel a distance 'b' to the screen S. In this region, there are no fields, so the particles move in a straight line with the velocity components $(v, v_y)$. The time taken to travel this distance 'b' is $t_b = \frac{b}{v}$. The additional transverse displacement ($y_b$) during this travel is:

$y_b = v_y t_b = \left(\frac{qEa}{mv}\right) \left(\frac{b}{v}\right) = \frac{qEab}{mv^2}$

4. Total shift on screen S:

The total shift $\Delta x$ on the screen S is the sum of the deflections from both parts:

$\Delta x = y_A + y_b = \frac{qEa^2}{2mv^2} + \frac{qEab}{mv^2}$

$\Delta x = \frac{qEa}{mv^2} \left(\frac{a}{2} + b\right)$

5. Substitute velocity and solve for specific charge:

Now, substitute the velocity $v = \frac{E}{B}$ (from Scenario 1) into the equation for $\Delta x$:

$\Delta x = \frac{qEa}{m(E/B)^2} \left(\frac{a}{2} + b\right)$

$\Delta x = \frac{qEaB^2}{mE^2} \left(\frac{a}{2} + b\right)$

$\Delta x = \frac{qB^2a}{mE} \left(\frac{a}{2} + b\right)$

Rearrange the equation to find the specific charge $\frac{q}{m}$:

$\frac{q}{m} = \frac{\Delta x E}{aB^2 \left(\frac{a}{2} + b\right)}$

To simplify the denominator:

$\frac{q}{m} = \frac{\Delta x E}{aB^2 \left(\frac{a+2b}{2}\right)}$

$\frac{q}{m} = \frac{2 \Delta x E}{aB^2 (a+2b)}$

The specific charge $q/m$ of the particles is $\frac{2 \Delta x E}{aB^2 (a+2b)}$.