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

To solve the problem, we need to find the specific charge \( \frac{q}{m} \) of the charged particles based on the given conditions. Let's break down the solution step by step.

### Step 1: Understanding the Initial Conditions
The charged particles move through a region with electric field \( E \) and magnetic field \( B \) without deviation. This means that the electric force is balanced by the magnetic force.

**Equation:**
\[ qE = qvB \]
Where:
- \( q \) is the charge of the particle,
- \( E \) is the electric field strength,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength.

### Step 2: Finding the Velocity
From the equation \( qE = qvB \), we can simplify to find the velocity \( v \):
\[ E = vB \implies v = \frac{E}{B} \]

### Step 3: Analyzing the Situation When the Magnetic Field is Switched Off
When the magnetic field is switched off, the particles will experience only the electric field, causing them to accelerate. The displacement \( \Delta x \) on the screen can be described using the equation of motion.

**Equation:**
\[ \Delta x = \frac{1}{2} a t^2 \]
Where \( a \) is the acceleration caused by the electric field \( E \):
\[ a = \frac{qE}{m} \]

### Step 4: Finding the Time of Flight
The time \( t \) it takes for the particles to travel through the distance \( a \) (the length of the region with fields) is given by:
\[ t = \frac{a}{v} = \frac{aB}{E} \]

### Step 5: Substituting Time into the Displacement Equation
Substituting \( t \) into the displacement equation:
\[ \Delta x = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{aB}{E}\right)^2 \]
This simplifies to:
\[ \Delta x = \frac{1}{2} \frac{qE}{m} \cdot \frac{a^2B^2}{E^2} \]
\[ \Delta x = \frac{1}{2} \frac{q a^2 B^2}{m E} \]

### Step 6: Rearranging to Find \( \frac{q}{m} \)
Rearranging the equation to isolate \( \frac{q}{m} \):
\[ \frac{q}{m} = \frac{2 \Delta x E}{a^2 B^2} \]

### Final Result
Thus, the specific charge \( \frac{q}{m} \) of the particles is given by:
\[ \frac{q}{m} = \frac{2 \Delta x E}{a^2 B^2} \]

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