Problem 32 - University Physics with Modern Physics

Question

In the Bainbridge mass spectrometer (see Fig. 27.24), the magnetic-field magnitude in the velocity selector is 0.510 T, and ions having a speed of 1.82 \(\times\) 10\(^6\) m/s pass through undeflected. (a) What is the electric-field magnitude in the velocity selector? (b) If the separation of the plates is 5.20 mm, what is the potential difference between the plates?

Step-by-Step Solution

Answer

(a) 9.28 × 10^5 V/m (b) 4828 V

Step 1: Analyze the equilibrium in the velocity selector

In the velocity selector, ions pass through undeflected, meaning the electric force and magnetic force acting on the ions are equal but in opposite directions. The electric force \(F_E\) is given by \(F_E = qE\), and the magnetic force \(F_B\) is given by \(F_B = qvB\). Therefore, at equilibrium condition, \(qE = qvB\).

Step 2: Solve for the electric field

From \(qE = qvB\), we can cancel the charge \(q\) on both sides because it appears in both terms. This simplifies to \(E = vB\), where \(E\) is the electric-field magnitude, \(v\) is the ion speed, and \(B\) is the magnetic-field magnitude. Substitute \(v = 1.82 \times 10^6 \, \text{m/s}\) and \(B = 0.510 \, \text{T}\) to find \(E\).\[E = (1.82 \times 10^6 \, \text{m/s})(0.510 \, \text{T})\]

Step 3: Calculate the electric field

Calculate the product from Step 2:\[E = (1.82 \times 10^6) \cdot (0.510) = 9.28 \times 10^5 \, \text{N/C or V/m}\]The electric-field magnitude in the velocity selector is \(9.28 \times 10^5 \, \text{V/m}\).

Step 4: Relate potential difference to electric field

The relationship between electric field \(E\), potential difference \(V\), and separation of the plates \(d\) is given by \(E = \frac{V}{d}\). Rearrange this formula to solve for the potential difference:\[V = Ed\]

Step 5: Calculate the potential difference

Substitute the given values into the formula from Step 4. Use the electric field \(E = 9.28 \times 10^5 \, \text{V/m}\) and separation \(d = 5.20 \, \text{mm} = 5.20 \times 10^{-3} \, \text{m}\).\[V = (9.28 \times 10^5 \, \text{V/m}) \cdot (5.20 \times 10^{-3} \, \text{m})\]Calculate the potential difference:\[V = 4828.32 \, \text{V}\]The potential difference between the plates is approximately \(4828 \, \text{V}\).

Key Concepts

Velocity Selector

In a Bainbridge mass spectrometer, the velocity selector is a crucial component that allows only ions with a specific velocity to pass through undeflected. This is achieved by carefully balancing electric and magnetic fields. When a charged particle, like an ion, moves through these fields at a certain speed, it experiences two forces:

• Magnetic force (\( F_B \)): Acts perpendicularly to its direction of motion, calculated as \( F_B = qvB \), where \( q \) is the charge, \( v \) is velocity, and \( B \) is the magnetic field strength.
• Electric force (\( F_E \)): Acts in the direction of the electric field, calculated as \( F_E = qE \), where \( E \) is the electric field strength.

For the ions to pass through the velocity selector, these forces must be equal in magnitude but opposite in direction. Therefore, setting \( F_E = F_B \) ensures that \( E = vB \), enabling only ions of the right speed to pass straight through without deflection.

Electric Field Calculation

Calculating the electric field in the velocity selector is essential for determining the conditions under which ions pass through undeflected. Once we know the ion's velocity and the magnetic field's strength, the electric field can be calculated using the relationship \( E = vB \). Here’s how it works:

• Identify the ion's velocity, \( v \), which is given as \( 1.82 \times 10^6 \, \text{m/s} \)
• Understand that the magnetic field strength is \( B = 0.510 \, \text{T} \)
• Substitute these values into the formula: \( E = (1.82 \times 10^6) \, \text{m/s} \times 0.510 \, \text{T} \)
• Compute to find \( E = 9.28 \times 10^5 \, \text{V/m} \)

The resulting electric field is strong enough to counterbalance the magnetic force experienced by ions of the specified velocity, allowing for precise mass spectrometry measurements.

Potential Difference

Understanding the relationship between electric field and potential difference is key when dealing with the separation of charged plates in a device like a velocity selector. The potential difference \( V \) between the plates is directly related to the electric field \( E \) and the separation distance \( d \) between them. The formula used is:\[E = \frac{V}{d}\]By rearranging this equation, we find:\[V = Ed\]To calculate the potential difference:

• Use the previously calculated electric field: \( E = 9.28 \times 10^5 \, \text{V/m} \)
• The plate separation is \( d = 5.20 \, \text{mm} = 5.20 \times 10^{-3} \, \text{m} \)
• Substitute into the formula to find \( V = (9.28 \times 10^5) \times (5.20 \times 10^{-3}) \)
• The result is approximately \( V = 4828 \, \text{V} \)

This potential difference ensures that the electric field is precisely tuned to the velocity of ions passing through the velocity selector, enabling their successful deflection and measurement.