The National High Magnetic Field Laboratory once held the world record for creating the strongest magnetic field. Their largest hybrid magnet can produce a constant magnetic field of 45 T. To see if such a strong magnetic field could pose health risks for nearby workers, calculate the maximum acceleration amax the field could produce for Na+ ions (of mass 3.8 x 10-26 kg) in blood traveling through the aorta. The speed of blood is highly variable, but 55 cm/s is reasonable in the aorta.
The National High Magnetic Field Laboratory once held the world record for creating the strongest magnetic field. Their largest hybrid magnet can produce a constant magnetic field of 45 T. To see if such a strong magnetic field could pose health risks for nearby workers, calculate the maximum acceleration amax the field could produce for Na+ ions (of mass 3.8 x 10-26 kg) in blood traveling through the aorta. The speed of blood is highly variable, but 55 cm/s is reasonable in the aorta.
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![**Title: Understanding the Effects of Strong Magnetic Fields on Ion Acceleration in the Human Body**
**Introduction:**
The National High Magnetic Field Laboratory once held the world record for creating the strongest magnetic field. Their largest hybrid magnet can produce a constant magnetic field of 45 Tesla (T).
**Objective:**
This exercise aims to assess whether such a strong magnetic field could pose health risks for nearby workers by calculating the maximum acceleration (\(a_{\text{max}}\)) that the field could produce for \( \text{Na}^+ \) ions (Sodium ions) in the blood traveling through the aorta.
**Given Data:**
- Magnetic Field Strength (B): 45 T
- Mass of \( \text{Na}^+ \) ions (\(m\)): \(3.8 \times 10^{-26} \) kg
- Speed of blood through the aorta (\(v\)): 55 cm/s (which is reasonable in the aorta but is highly variable)
**Calculation:**
To calculate the maximum acceleration (\( a_{\text{max}} \)) experienced by the \( \text{Na}^+ \) ions in the magnetic field, use the following formula for magnetic force \(F_B\):
\[ F_B = qvB \]
Where:
- \( q \) is the charge of the ion (for \( \text{Na}^+ \), \( q = +1e \), where \( e = 1.6 \times 10^{-19} \) Coulombs)
- \( v \) is the speed of the ion (55 cm/s, converted to meters: 0.55 m/s)
- \( B \) is the magnetic field (45 T)
Since \( F = ma \),
\[ a_{\text{max}} = \frac{F}{m} = \frac{qvB}{m} \]
Fill in the values to find \( a_{\text{max}} \):
\[ a_{\text{max}} = \frac{(1.6 \times 10^{-19} \text{ C})(0.55 \text{ m/s})(45 \text{ T})}{3.8 \times 10^{-26} \text{ kg}} \]
**Result:**
\[ a_{\text{max}} = \boxed{ \hspace{40pt} } \text{ m/s}^2 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37e1db01-06d0-4e20-aae1-26f96b6814c3%2F77dd1097-745a-4549-9b5c-f6c07bf6acaf%2Fwx5w0am_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Understanding the Effects of Strong Magnetic Fields on Ion Acceleration in the Human Body**
**Introduction:**
The National High Magnetic Field Laboratory once held the world record for creating the strongest magnetic field. Their largest hybrid magnet can produce a constant magnetic field of 45 Tesla (T).
**Objective:**
This exercise aims to assess whether such a strong magnetic field could pose health risks for nearby workers by calculating the maximum acceleration (\(a_{\text{max}}\)) that the field could produce for \( \text{Na}^+ \) ions (Sodium ions) in the blood traveling through the aorta.
**Given Data:**
- Magnetic Field Strength (B): 45 T
- Mass of \( \text{Na}^+ \) ions (\(m\)): \(3.8 \times 10^{-26} \) kg
- Speed of blood through the aorta (\(v\)): 55 cm/s (which is reasonable in the aorta but is highly variable)
**Calculation:**
To calculate the maximum acceleration (\( a_{\text{max}} \)) experienced by the \( \text{Na}^+ \) ions in the magnetic field, use the following formula for magnetic force \(F_B\):
\[ F_B = qvB \]
Where:
- \( q \) is the charge of the ion (for \( \text{Na}^+ \), \( q = +1e \), where \( e = 1.6 \times 10^{-19} \) Coulombs)
- \( v \) is the speed of the ion (55 cm/s, converted to meters: 0.55 m/s)
- \( B \) is the magnetic field (45 T)
Since \( F = ma \),
\[ a_{\text{max}} = \frac{F}{m} = \frac{qvB}{m} \]
Fill in the values to find \( a_{\text{max}} \):
\[ a_{\text{max}} = \frac{(1.6 \times 10^{-19} \text{ C})(0.55 \text{ m/s})(45 \text{ T})}{3.8 \times 10^{-26} \text{ kg}} \]
**Result:**
\[ a_{\text{max}} = \boxed{ \hspace{40pt} } \text{ m/s}^2 \]
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