Answer :
Certainly! Let's work step-by-step to find out the magnetic induction at which the orientation energy is comparable to the thermal energy at room temperature.
### Step 1: Understand the Given Data
- Magnetic moment (\(\mu_m\)):
[tex]\[ \mu_m = 5 \times \mu_B = 5 \times 9.274009994 \times 10^{-24} \, \text{J/T} \][/tex]
where \(\mu_B\) (Bohr magneton) is \(9.274009994 \times 10^{-24} \, \text{J/T}\).
- Boltzmann constant (\(k_B\)):
[tex]\[ k_B = 1.380649 \times 10^{-23} \, \text{J/K} \][/tex]
- Room temperature (\(T\)):
[tex]\[ T = 300 \, \text{K} \][/tex]
### Step 2: Calculate Thermal Energy
The thermal energy at room temperature can be calculated using:
[tex]\[ E_{\text{thermal}} = k_B \times T \][/tex]
Substituting the values:
[tex]\[ E_{\text{thermal}} = 1.380649 \times 10^{-23} \, \text{J/K} \times 300 \, \text{K} = 4.141947 \times 10^{-21} \, \text{J} \][/tex]
### Step 3: Equating Orientation Energy to Thermal Energy
The orientation energy in a magnetic field (\(B\)) is given by:
[tex]\[ E_{\text{orientation}} = \mu_m \times B \][/tex]
To find the magnetic induction (\(B\)) where orientation energy is comparable to the thermal energy, set \(E_{\text{orientation}} = E_{\text{thermal}}\):
[tex]\[ \mu_m \times B = E_{\text{thermal}} \][/tex]
### Step 4: Solve for Magnetic Induction (\(B\))
Re-arranging the above formula to solve for \(B\):
[tex]\[ B = \frac{E_{\text{thermal}}}{\mu_m} \][/tex]
Substitute the values:
[tex]\[ \mu_m = 5 \times 9.274009994 \times 10^{-24} \, \text{J/T} = 4.637004997 \times 10^{-23} \, \text{J/T} \][/tex]
[tex]\[ B = \frac{4.141947 \times 10^{-21} \, \text{J}}{4.637004997 \times 10^{-23} \, \text{J/T}} \][/tex]
By performing the division:
[tex]\[ B \approx 89.32 \, \text{T} \][/tex]
### Conclusion
The magnetic induction ([tex]\(B\)[/tex]) must be approximately [tex]\(89.32\)[/tex] Tesla for the orientation energy to be comparable to the thermal energy at room temperature.
### Step 1: Understand the Given Data
- Magnetic moment (\(\mu_m\)):
[tex]\[ \mu_m = 5 \times \mu_B = 5 \times 9.274009994 \times 10^{-24} \, \text{J/T} \][/tex]
where \(\mu_B\) (Bohr magneton) is \(9.274009994 \times 10^{-24} \, \text{J/T}\).
- Boltzmann constant (\(k_B\)):
[tex]\[ k_B = 1.380649 \times 10^{-23} \, \text{J/K} \][/tex]
- Room temperature (\(T\)):
[tex]\[ T = 300 \, \text{K} \][/tex]
### Step 2: Calculate Thermal Energy
The thermal energy at room temperature can be calculated using:
[tex]\[ E_{\text{thermal}} = k_B \times T \][/tex]
Substituting the values:
[tex]\[ E_{\text{thermal}} = 1.380649 \times 10^{-23} \, \text{J/K} \times 300 \, \text{K} = 4.141947 \times 10^{-21} \, \text{J} \][/tex]
### Step 3: Equating Orientation Energy to Thermal Energy
The orientation energy in a magnetic field (\(B\)) is given by:
[tex]\[ E_{\text{orientation}} = \mu_m \times B \][/tex]
To find the magnetic induction (\(B\)) where orientation energy is comparable to the thermal energy, set \(E_{\text{orientation}} = E_{\text{thermal}}\):
[tex]\[ \mu_m \times B = E_{\text{thermal}} \][/tex]
### Step 4: Solve for Magnetic Induction (\(B\))
Re-arranging the above formula to solve for \(B\):
[tex]\[ B = \frac{E_{\text{thermal}}}{\mu_m} \][/tex]
Substitute the values:
[tex]\[ \mu_m = 5 \times 9.274009994 \times 10^{-24} \, \text{J/T} = 4.637004997 \times 10^{-23} \, \text{J/T} \][/tex]
[tex]\[ B = \frac{4.141947 \times 10^{-21} \, \text{J}}{4.637004997 \times 10^{-23} \, \text{J/T}} \][/tex]
By performing the division:
[tex]\[ B \approx 89.32 \, \text{T} \][/tex]
### Conclusion
The magnetic induction ([tex]\(B\)[/tex]) must be approximately [tex]\(89.32\)[/tex] Tesla for the orientation energy to be comparable to the thermal energy at room temperature.
