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Your homework involves finding the smallest value of the magnetic field \( B \) that will keep an aluminum ion trapped inside a square-shaped magnetic field region with the following properties:

- Side length of the square: \( L = 85 \, \text{cm} = 0.85 \, \text{m} \)
- Mass of the aluminum ion: \( m = 4.5 \times 10^{-26} \, \text{kg} \)
- Charge of the aluminum ion: \( q = 4.8 \times 10^{-19} \, \text{C} \)
- Velocity of the ion: \( v = 350 \, \text{m/s} \)

### Key Physics Concepts:
The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force, which acts as the centripetal force when the particle is in circular motion. For the ion to stay inside the magnetic field region, its radius of motion must be smaller than or equal to half the side length of the square region.

1. **Magnetic Force and Circular Motion:**
   The magnetic force provides the centripetal force for circular motion:
   \[
   F_{\text{mag}} = qvB = \frac{mv^2}{r}
   \]
   Solving for \( r \):
   \[
   r = \frac{mv}{qB}
   \]

2. **Maximum Allowable Radius:**
   The ion must stay inside the square region, so the radius of its path must not exceed half of the side length:
   \[
   r \leq \frac{L}{2}
   \]
   Combining this with the equation for \( r \), we get:
   \[
   \frac{mv}{qB} \leq \frac{L}{2}
   \]
   Solving for \( B \):
   \[
   B \geq \frac{2mv}{qL}
   \]

### Symbolic Answer:
The smallest magnetic field \( B \) required to keep the ion trapped is:
\[
B = \frac{2mv}{qL}
\]

### Plugging in the values:
- \( m = 4.5 \times 10^{-26} \, \text{kg} \)
- \( v = 350 \, \text{m/s} \)
- \( q = 4.8 \times 10^{-19} \, \text{C} \)
- \( L = 0.85 \, \text{m} \)

\[
B = \frac{2 \times (4.5 \times 10^{-26}) \times 350}{(4.8 \times 10^{-19}) \times 0.85}
\]
Calculating this:
\[
B \approx 7.65 \times 10^{-5} \, \text{T}
\]

### Final Answer:
The smallest value of the magnetic field \( B \) is approximately \( 7.65 \times 10^{-5} \, \text{T} \).

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Let me know if you need more details or have any questions. Here are 5 related questions you might consider:

1. How would the result change if the speed of the ion doubled?
2. What would happen if the charge of the ion were negative?
3. How does the size of the magnetic field region affect the ion’s motion?
4. What happens if the magnetic field is not uniform?
5. How would you experimentally determine the magnetic field strength?

**Tip:** Always check that units are consistent before calculating the final numerical value!

Learn how to find the smallest value of the magnetic field required to trap an aluminum ion in a square-shaped magnetic field region. This problem uses key physics concepts such as magnetic force and centripetal motion. The ion has a mass of 4.5 × 10^-26 kg, charge of 4.8 × 10^-19 C, and velocity of 350 m/s within a square region with side length 85 cm.

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