Which of the following particles will describe will experience maximum magnetic force(magnitude) when projected with the same velocity perpendicular to a magnetic field?

Options

A. Electron
B. Proton
C. `He^+`
D. `Li^(++)`

Answer

To determine which particle will experience the maximum magnetic force when projected with the same velocity perpendicular to a magnetic field, we can use the formula for magnetic force:

\[ F = q \cdot v \cdot B \cdot \sin(\theta) \]

Where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

Since the particles are projected perpendicular to the magnetic field, \( \sin(90^\circ) = 1 \), and the formula simplifies to:

\[ F = q \cdot v \cdot B \]

Given that the velocity \( v \) and the magnetic field \( B \) are the same for all particles, the magnetic force \( F \) will depend solely on the charge \( q \) of each particle.

Now, let's analyze the charge of each particle:

1. **Electron**:
- Charge \( q = -1.6 \times 10^{-19} \) C (magnitude = \( 1.6 \times 10^{-19} \) C)

2. **Proton**:
- Charge \( q = +1.6 \times 10^{-19} \) C (magnitude = \( 1.6 \times 10^{-19} \) C)

3. **Helium ion (He\(^+\))**:
- Charge \( q = +2 \times 1.6 \times 10^{-19} \) C (magnitude = \( 3.2 \times 10^{-19} \) C)

4. **Lithium ion (Li\(^{2+}\))**:
- Charge \( q = +3 \times 1.6 \times 10^{-19} \) C (magnitude = \( 4.8 \times 10^{-19} \) C)

Now, we can summarize the magnitudes of the charges:
- Electron: \( 1.6 \times 10^{-19} \) C
- Proton: \( 1.6 \times 10^{-19} \) C
- Helium ion (He\(^+\)): \( 3.2 \times 10^{-19} \) C
- Lithium ion (Li\(^{2+}\)): \( 4.8 \times 10^{-19} \) C

From the above analysis, we can see that the lithium ion (Li\(^{2+}\)) has the highest charge magnitude.

Thus, the particle that will experience the maximum magnetic force is:

**Lithium ion (Li\(^{2+}\))**.