Edits to “Statement”, “Answer”
en/9.2.21.md
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| ### Statement | |||
| − | $9.2.21.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $9.2.21.$ Estimate the magnetic field induction at the center of a flat iron ring of thickness $1$ cm with an internal radius of $10$ cm and an external radius of $20$ cm. All iron atoms are oriented along the axis of the ring, the magnetic moment of an iron atom is $2\mu_e = 1.85\cdot10^{-23}$ J/T. | |||
| ### Solution | |||
| We will use the solution to the previous problem (see [9.2.20](https://savchenkosolutions.com/en/9.2.20)). The field at short distances can be expressed through the result of the previous problem as: | |||
| $$ | |||
| B = B_{R_1} - B_{R_2} = \frac{\mu_0 M h}{2} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) | |||
| $$ | |||
| Let's estimate the magnetic moment per unit volume (magnetization): | |||
| $$ | |||
| M = \frac{2\rho N_A \mu_e}{M_{Fe}} | |||
| $$ | |||
| Substituting and calculating: | |||
| $$ | |||
| B = \frac{\mu_0\rho N_A \mu_eh}{M_{Fe}} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) \approx 4.9 \cdot 10^{-2} \text{ T} | |||
| $$ | |||
| #### Answer | |||
| @@ -33,7 +27,3 @@Answer | |||
| $$ | |||
| B = \frac{\mu_0\rho N_A \mu_eh}{M_{Fe}} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) \approx 4.9 \cdot 10^{-2} \text{ T} | |||
| $$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $9.2.21.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $9.2.21.$ Estimate the magnetic field induction at the center of a flat iron ring of thickness $1$ cm with an internal radius of $10$ cm and an external radius of $20$ cm. All iron atoms are oriented along the axis of the ring, the magnetic moment of an iron atom is $2\mu_e = 1.85\cdot10^{-23}$ J/T. | $9.2.21.$ Estimate the magnetic field induction at the center of a flat iron ring of thickness $1$ cm with an internal radius of $10$ cm and an external radius of $20$ cm. All iron atoms are oriented along the axis of the ring, the magnetic moment of an iron atom is $2\mu_e = 1.85\cdot10^{-23}$ J/T. | ||
| ### Solution | ### Solution | ||
| We will use the solution to the previous problem (see [9.2.20](https://savchenkosolutions.com/en/9.2.20)). The field at short distances can be expressed through the result of the previous problem as: | We will use the solution to the previous problem (see [9.2.20](https://savchenkosolutions.com/en/9.2.20)). The field at short distances can be expressed through the result of the previous problem as: | ||
| $$ | $$ | ||
| B = B_{R_1} - B_{R_2} = \frac{\mu_0 M h}{2} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) | B = B_{R_1} - B_{R_2} = \frac{\mu_0 M h}{2} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) | ||
| $$ | $$ | ||
| Let's estimate the magnetic moment per unit volume (magnetization): | Let's estimate the magnetic moment per unit volume (magnetization): | ||
| $$ | $$ | ||
| M = \frac{2\rho N_A \mu_e}{M_{Fe}} | M = \frac{2\rho N_A \mu_e}{M_{Fe}} | ||
| $$ | $$ | ||
| Substituting and calculating: | Substituting and calculating: | ||
| $$ | $$ | ||
| B = \frac{\mu_0\rho N_A \mu_eh}{M_{Fe}} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) \approx 4.9 \cdot 10^{-2} \text{ T} | B = \frac{\mu_0\rho N_A \mu_eh}{M_{Fe}} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) \approx 4.9 \cdot 10^{-2} \text{ T} | ||
| $$ | $$ | ||
| #### Answer | #### Answer | ||
| @@ -33,7 +27,3 @@Answer | |||
| $$ | $$ | ||
| B = \frac{\mu_0\rho N_A \mu_eh}{M_{Fe}} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) \approx 4.9 \cdot 10^{-2} \text{ T} | B = \frac{\mu_0\rho N_A \mu_eh}{M_{Fe}} \left( \frac{1}{R_1}-\frac{1}{R_2} \right) \approx 4.9 \cdot 10^{-2} \text{ T} | ||
| $$ | $$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||