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Sagot :
To find the magnitude of the magnetic force exerted on the particle, we need to follow these steps:
### Step 1: Calculate the Kinetic Energy Gained by the Particle
When a particle is accelerated from rest through a potential difference, the kinetic energy it gains is given by:
[tex]\[ K.E. = q \cdot V \][/tex]
where:
- [tex]\( q \)[/tex] is the charge of the particle.
- [tex]\( V \)[/tex] is the potential difference.
Here, the charge [tex]\( q = 3.20 \times 10^{-19} \, \text{C} \)[/tex] and the potential difference [tex]\( V = 1.6 \times 10^7 \, \text{V} \)[/tex].
[tex]\[ K.E. = (3.20 \times 10^{-19} \, \text{C}) \times (1.6 \times 10^7 \, \text{V}) = 5.12 \times 10^{-12} \, \text{J} \][/tex]
### Step 2: Calculate the Velocity of the Particle
The kinetic energy of a particle moving with velocity [tex]\( v \)[/tex] and mass [tex]\( m \)[/tex] is given by:
[tex]\[ K.E. = \frac{1}{2} m v^2 \][/tex]
Solving for [tex]\( v \)[/tex], we get:
[tex]\[ v = \sqrt{\frac{2 \, K.E.}{m}} \][/tex]
Using the given mass [tex]\( m = 6.4 \times 10^{-27} \, \text{kg} \)[/tex] and kinetic energy [tex]\( K.E. = 5.12 \times 10^{-12} \, \text{J} \)[/tex],
[tex]\[ v = \sqrt{\frac{2 \times 5.12 \times 10^{-12} \, \text{J}}{6.4 \times 10^{-27} \, \text{kg}}} = \sqrt{1.6 \times 10^{15}} = 4.00 \times 10^7 \, \text{m/s} \][/tex]
### Step 3: Calculate the Magnetic Force
The magnetic force [tex]\( F \)[/tex] on a particle moving perpendicular to a magnetic field is given by:
[tex]\[ F = q \cdot v \cdot B \][/tex]
where:
- [tex]\( q \)[/tex] is the charge of the particle.
- [tex]\( v \)[/tex] is the velocity of the particle.
- [tex]\( B \)[/tex] is the magnetic field strength.
Using [tex]\( q = 3.20 \times 10^{-19} \, \text{C} \)[/tex], [tex]\( v = 4.00 \times 10^7 \, \text{m/s} \)[/tex], and [tex]\( B = 3.2 \, \text{T} \)[/tex],
[tex]\[ F = (3.20 \times 10^{-19} \, \text{C}) \times (4.00 \times 10^7 \, \text{m/s}) \times (3.2 \, \text{T}) = 4.096 \times 10^{-11} \, \text{N} \][/tex]
### Conclusion
The magnitude of the magnetic force exerted on the particle is:
[tex]\[ \boxed{4.09 \times 10^{-11} \, \text{N}} \][/tex]
This corresponds to the provided option [tex]\( 4.09 \times 10^{-11} \, \text{N} \)[/tex].
### Step 1: Calculate the Kinetic Energy Gained by the Particle
When a particle is accelerated from rest through a potential difference, the kinetic energy it gains is given by:
[tex]\[ K.E. = q \cdot V \][/tex]
where:
- [tex]\( q \)[/tex] is the charge of the particle.
- [tex]\( V \)[/tex] is the potential difference.
Here, the charge [tex]\( q = 3.20 \times 10^{-19} \, \text{C} \)[/tex] and the potential difference [tex]\( V = 1.6 \times 10^7 \, \text{V} \)[/tex].
[tex]\[ K.E. = (3.20 \times 10^{-19} \, \text{C}) \times (1.6 \times 10^7 \, \text{V}) = 5.12 \times 10^{-12} \, \text{J} \][/tex]
### Step 2: Calculate the Velocity of the Particle
The kinetic energy of a particle moving with velocity [tex]\( v \)[/tex] and mass [tex]\( m \)[/tex] is given by:
[tex]\[ K.E. = \frac{1}{2} m v^2 \][/tex]
Solving for [tex]\( v \)[/tex], we get:
[tex]\[ v = \sqrt{\frac{2 \, K.E.}{m}} \][/tex]
Using the given mass [tex]\( m = 6.4 \times 10^{-27} \, \text{kg} \)[/tex] and kinetic energy [tex]\( K.E. = 5.12 \times 10^{-12} \, \text{J} \)[/tex],
[tex]\[ v = \sqrt{\frac{2 \times 5.12 \times 10^{-12} \, \text{J}}{6.4 \times 10^{-27} \, \text{kg}}} = \sqrt{1.6 \times 10^{15}} = 4.00 \times 10^7 \, \text{m/s} \][/tex]
### Step 3: Calculate the Magnetic Force
The magnetic force [tex]\( F \)[/tex] on a particle moving perpendicular to a magnetic field is given by:
[tex]\[ F = q \cdot v \cdot B \][/tex]
where:
- [tex]\( q \)[/tex] is the charge of the particle.
- [tex]\( v \)[/tex] is the velocity of the particle.
- [tex]\( B \)[/tex] is the magnetic field strength.
Using [tex]\( q = 3.20 \times 10^{-19} \, \text{C} \)[/tex], [tex]\( v = 4.00 \times 10^7 \, \text{m/s} \)[/tex], and [tex]\( B = 3.2 \, \text{T} \)[/tex],
[tex]\[ F = (3.20 \times 10^{-19} \, \text{C}) \times (4.00 \times 10^7 \, \text{m/s}) \times (3.2 \, \text{T}) = 4.096 \times 10^{-11} \, \text{N} \][/tex]
### Conclusion
The magnitude of the magnetic force exerted on the particle is:
[tex]\[ \boxed{4.09 \times 10^{-11} \, \text{N}} \][/tex]
This corresponds to the provided option [tex]\( 4.09 \times 10^{-11} \, \text{N} \)[/tex].
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