Let's tackle the problem part by part. --- ### Given: - **Number of particles:** \( N \) - **Magnetic moment per particle:** \( \mu \) - **Two energy levels:** \( +\mu x_0/2 \) (upper, antiparallel), \( -\mu x_0/2 \) (lower, parallel) when the magnetic intensity is \( x_0 \). - **System prepared so that 1/3 of particles are in the upper level, 2/3 in the lower level, and isolated.** Let's denote: - Number in upper level: \( N_u = N/3 \) - Number in lower level: \( N_l = 2N/3 \) --- ## (a) **Find the energy and net magnetic moment of the system.** **Energies:** - Upper level: \( +\mu x_0/2 \) - Lower level: \( -\mu x_0/2 \) ### **Total Energy** \[ E = N_u \left( +\frac{\mu x_0}{2} \right) + N_l \left( -\frac{\mu x_0}{2} \right) \] Insert \( N_u = N/3 \), \( N_l = 2N/3 \): \[ E = \frac{N}{3} \left( +\frac{\mu x_0}{2} \right) + \frac{2N}{3} \left( -\frac{\mu x_0}{2} \right) \] \[ E = \frac{N \mu x_0}{6} - \frac{2N \mu x_0}{6} \] \[ E = -\frac{N \mu x_0}{6} \] --- ### **Net Magnetic Moment** - Upper level (antiparallel): Moment = \( -\mu \) - Lower level (parallel): Moment = \( +\mu \) \[ M = N_u (-\mu) + N_l (+\mu) \] \[ M = \frac{N}{3}(-\mu) + \frac{2N}{3}(+\mu) \] \[ M = -\frac{N\mu}{3} + \frac{2N\mu}{3} \] \[ M = \frac{N\mu}{3} \] --- ### **(a) Final Answers:** \[ \boxed{ E = -\frac{N \mu x_0}{6} } \] \[ \boxed{ M = \frac{N\mu}{3} } \] --- ## (b) **Change in energy and net magnetic moment when \( x_0 \) is reduced to \( x_0/2 \) (system isolated, no particle redistribution).** **Since the system is isolated and the populations don't change:** - \( N_u = N/3 \) - \( N_l = 2N/3 \) New energies: - Upper: \( +\frac{\mu}{2} \cdot \frac{x_0}{2} = +\frac{\mu x_0}{4} \) - Lower: \( -\frac{\mu x_0}{4} \) **New energy:** \[ E' = \frac{N}{3}\left(+\frac{\mu x_0}{4}\right) + \frac{2N}{3}\left(-\frac{\mu x_0}{4}\right) \] \[ E' = \frac{N\mu x_0}{12} - \frac{2N\mu x_0}{12} \] \[ E' = -\frac{N\mu x_0}{12} \] **Change in energy:** \[ \Delta E = E' - E = \left(-\frac{N\mu x_0}{12}\right) - \left(-\frac{N\mu x_0}{6}\right) \] \[ = -\frac{N\mu x_0}{12} + \frac{N\mu x_0}{6} \] \[ = \frac{N\mu x_0}{12} \] **Net magnetic moment does not change (no redistribution):** \[ M' = \frac{N\mu}{3} \] \[ \Delta M = M' - M = 0 \] --- ### **(b) Final Answers:** \[ \boxed{ \Delta E = \frac{N\mu x_0}{12} } \] \[ \boxed{ \Delta M = 0 } \] --- ## (c) **Change in net magnetic moment when \( x_0 \) is reduced to \( x_0/2 \) but the energy remains constant (system can exchange energy, particle distribution can change).** - **Energy remains \( E = -\frac{N\mu x_0}{6} \)** - **New energy levels: \( +\frac{\mu x_0}{4} \), \( -\frac{\mu x_0}{4} \)** - Let fraction in upper = \( f \), lower = \( 1-f \). \[ E = N \left[ f \left(+\frac{\mu x_0}{4}\right) + (1-f)\left(-\frac{\mu x_0}{4}\right) \right] \] \[ E = N\left[\frac{\mu x_0}{4}(f - (1-f))\right] \] \[ E = N\left[\frac{\mu x_0}{4}(2f-1)\right] \] Set equal to original energy: \[ N\left[\frac{\mu x_0}{4}(2f-1)\right] = -\frac{N\mu x_0}{6} \] \[ \frac{\mu x_0}{4}(2f-1) = -\frac{\mu x_0}{6} \] \[ 2f-1 = -\frac{2}{3} \] \[ 2f = 1 - \frac{2}{3} \] \[ 2f = \frac{1}{3} \] \[ f = \frac{1}{6} \] So, new distribution: - Upper: \( N/6 \) - Lower: \( 5N/6 \) **Net magnetic moment:** \[ M'' = N_u''(-\mu) + N_l''(+\mu) \] \[ M'' = \frac{N}{6}(-\mu) + \frac{5N}{6}(+\mu) \] \[ M'' = -\frac{N\mu}{6} + \frac{5N\mu}{6} \] \[ M'' = \frac{4N\mu}{6} = \frac{2N\mu}{3} \] **Change in net magnetic moment:** \[ \Delta M = M'' - M = \frac{2N\mu}{3} - \frac{N\mu}{3} = \frac{N\mu}{3} \] --- ### **(c) Final Answers:** \[ \boxed{ \Delta M = \frac{N\mu}{3} } \] --- ## **Summary Table** | Part | Energy (\(E\)) | Net Magnetic Moment (\(M\)) | Change in Energy (\(\Delta E\)) | Change in Magnetic Moment (\(\Delta M\)) | |------|----------------|-----------------------------|-----------------------------------|------------------------------------------| | (a) | \( -\frac{N\mu x_0}{6} \) | \( \frac{N\mu}{3} \) | - | - | | (b) | \( -\frac{N\mu x_0}{12} \) | \( \frac{N\mu}{3} \) | \( \frac{N\mu x_0}{12} \) | 0 | | (c) | \( -\frac{N\mu x_0}{6} \) (constant) | \( \frac{2N\mu}{3} \) | - | \( \frac{N\mu}{3} \) | Let me know if you need further explanation for any step!

**Final Answers:** **(a)** Energy: \(\boxed{E = -\frac{N \mu x_0}{6}}\) Net magnetic moment: \(\boxed{M = \frac{N \mu}{3}}\) **(b)** Change in energy: \(\boxed{\Delta E = \frac{N \mu x_0}{12}}\) Change in magnetic moment: \(\boxed{\Delta M = 0}\) **(c)** Change in magnetic moment: \(\boxed{\Delta M = \frac{N \mu}{3}}\)