Question
What is the magnetic force required to keep a deuteron moving in a circular path in a cyclotron?
Answers
GPT-4o mini
To calculate the magnetic force required to keep a deuteron moving in a circular path in a cyclotron, we can use the formula for the magnetic force acting on a charged particle moving in a magnetic field. This force must be equal to the centripetal force needed to keep the particle moving along a circular path.
1. **Magnetic Force**: The magnetic force \( F \) on a charged particle is given by the equation:
\[
F = qvB
\]
where:
- \( q \) = charge of the particle,
- \( v \) = velocity of the particle,
- \( B \) = magnetic field strength.
2. **Centripetal Force**: The centripetal force \( F_c \) required to keep an object moving in a circle is given by:
\[
F_c = \frac{mv^2}{r}
\]
where:
- \( m \) = mass of the particle,
- \( r \) = radius of the circular path.
In a cyclotron, the magnetic force provides the necessary centripetal force to keep the particle in a circular motion, so we set the magnetic force equal to the centripetal force:
\[
qvB = \frac{mv^2}{r}
\]
3. **Solving for the magnetic field**: We can rearrange this equation to express the magnetic field \( B \):
\[
B = \frac{mv}{qr}
\]
### Deuteron Properties
- The charge \( q \) of a deuteron (which is a nucleus of deuterium, consisting of one proton and one neutron) is equal to the charge of a proton: \( q \approx 1.6 \times 10^{-19} \) Coulombs.
- The mass \( m \) of a deuteron is approximately \( 2 \times 1.67 \times 10^{-27} \) kg = \( 3.34 \times 10^{-27} \) kg (where 1 amu is approximately \( 1.67 \times 10^{-27} \) kg).
### Example Calculation
Assuming you have values for velocity \( v \) and radius \( r \), you can substitute those into the rearranged equation:
For example, if:
- \( v = 2 \times 10^6 \) m/s (a typical value in a cyclotron),
- \( r = 0.5 \) m,
We plug in the values:
\[
B = \frac{(3.34 \times 10^{-27} \text{ kg})(2 \times 10^6 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C})(0.5 \text{ m})}
\]
Calculating this gives:
\[
B = \frac{6.68 \times 10^{-21}}{0.8 \times 10^{-19}} = 0.0835 \text{ T} = 83.5 \text{ mT}
\]
Hence, the required magnetic field \( B \) is \( 83.5 \) mT in this example.
### Conclusion
The magnetic force required to keep a deuteron moving in a circular path in a cyclotron is equal to the centripetal force needed for that motion, effectively determined by \( qvB \) for the charge, velocity, and magnetic field strength involved in the cyclotron.
1. **Magnetic Force**: The magnetic force \( F \) on a charged particle is given by the equation:
\[
F = qvB
\]
where:
- \( q \) = charge of the particle,
- \( v \) = velocity of the particle,
- \( B \) = magnetic field strength.
2. **Centripetal Force**: The centripetal force \( F_c \) required to keep an object moving in a circle is given by:
\[
F_c = \frac{mv^2}{r}
\]
where:
- \( m \) = mass of the particle,
- \( r \) = radius of the circular path.
In a cyclotron, the magnetic force provides the necessary centripetal force to keep the particle in a circular motion, so we set the magnetic force equal to the centripetal force:
\[
qvB = \frac{mv^2}{r}
\]
3. **Solving for the magnetic field**: We can rearrange this equation to express the magnetic field \( B \):
\[
B = \frac{mv}{qr}
\]
### Deuteron Properties
- The charge \( q \) of a deuteron (which is a nucleus of deuterium, consisting of one proton and one neutron) is equal to the charge of a proton: \( q \approx 1.6 \times 10^{-19} \) Coulombs.
- The mass \( m \) of a deuteron is approximately \( 2 \times 1.67 \times 10^{-27} \) kg = \( 3.34 \times 10^{-27} \) kg (where 1 amu is approximately \( 1.67 \times 10^{-27} \) kg).
### Example Calculation
Assuming you have values for velocity \( v \) and radius \( r \), you can substitute those into the rearranged equation:
For example, if:
- \( v = 2 \times 10^6 \) m/s (a typical value in a cyclotron),
- \( r = 0.5 \) m,
We plug in the values:
\[
B = \frac{(3.34 \times 10^{-27} \text{ kg})(2 \times 10^6 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C})(0.5 \text{ m})}
\]
Calculating this gives:
\[
B = \frac{6.68 \times 10^{-21}}{0.8 \times 10^{-19}} = 0.0835 \text{ T} = 83.5 \text{ mT}
\]
Hence, the required magnetic field \( B \) is \( 83.5 \) mT in this example.
### Conclusion
The magnetic force required to keep a deuteron moving in a circular path in a cyclotron is equal to the centripetal force needed for that motion, effectively determined by \( qvB \) for the charge, velocity, and magnetic field strength involved in the cyclotron.