Let's solve the problem **step by step** and explain each step: --- ### **Given Data** - Wavelength of probe light, \( \lambda_0 = 0.53 \, \mu m = 0.53 \times 10^{-6} \) m - Magnetic field, \( B = 100 \) T - Electron density, \( n = 10^{27} \) m\(^{-3}\) - Path length, \( L = 30 \, \mu m = 30 \times 10^{-6} \) m - Assume \( \vec{k} \parallel \vec{B} \) --- ### **Step 1: Write the Faraday Rotation Formula** The Faraday rotation angle \( \theta \) (in radians) in a plasma is given by: \[ \theta = \frac{e^3 \lambda_0^2}{8 \pi^2 \epsilon_0 m_e^2 c^3} n B L \] #### **Explanation:** - This formula relates the rotation angle of a plane-polarized light as it propagates through a plasma in the presence of a magnetic field parallel to the direction of light. - \(e\) is the elementary charge, \(m_e\) is the electron mass, \(\epsilon_0\) is the vacuum permittivity, \(c\) is the speed of light. --- ### **Step 2: Insert Known Constants** - \( e = 1.602 \times 10^{-19} \) C - \( m_e = 9.109 \times 10^{-31} \) kg - \( \epsilon_0 = 8.854 \times 10^{-12} \) F/m - \( c = 3.00 \times 10^8 \) m/s --- ### **Step 3: Plug in Values** \[ \theta = \frac{(1.602 \times 10^{-19})^3 \times (0.53 \times 10^{-6})^2}{8 \pi^2 \times 8.854 \times 10^{-12} \times (9.109 \times 10^{-31})^2 \times (3.00 \times 10^8)^3} \times (10^{27}) \times (100) \times (30 \times 10^{-6}) \] --- ### **Step 4: Calculate Numerator and Denominator Separately** #### **Numerator:** \[ (1.602 \times 10^{-19})^3 = 4.108 \times 10^{-57} \] \[ (0.53 \times 10^{-6})^2 = 0.2809 \times 10^{-12} = 2.809 \times 10^{-13} \] \[ \text{So, Numerator} = 4.108 \times 10^{-57} \times 2.809 \times 10^{-13} = 1.154 \times 10^{-69} \] #### **Denominator:** \[ 8 \pi^2 = 8 \times 9.8696 = 78.9568 \] \[ (9.109 \times 10^{-31})^2 = 8.299 \times 10^{-61} \] \[ (3.00 \times 10^8)^3 = 2.7 \times 10^{25} \] \[ \text{So, Denominator} = 78.9568 \times 8.854 \times 10^{-12} \times 8.299 \times 10^{-61} \times 2.7 \times 10^{25} \] Calculate step by step: - \( 78.9568 \times 8.854 \times 10^{-12} = 698.775 \times 10^{-12} = 6.98775 \times 10^{-10} \) - \( 6.98775 \times 10^{-10} \times 8.299 \times 10^{-61} = 57.973 \times 10^{-71} = 5.7973 \times 10^{-70} \) - \( 5.7973 \times 10^{-70} \times 2.7 \times 10^{25} = 15.6527 \times 10^{-45} = 1.565 \times 10^{-44} \) --- ### **Step 5: Evaluate the Fraction** Now, \[ \theta = \frac{1.154 \times 10^{-69}}{1.565 \times 10^{-44}} \times (10^{27}) \times (100) \times (30 \times 10^{-6}) \] \[ = (7.375 \times 10^{-26}) \times (10^{27}) \times (100) \times (30 \times 10^{-6}) \] \[ = (7.375 \times 10^1) \times (30 \times 10^{-6}) \] \[ = 73.75 \times 30 \times 10^{-6} \] \[ = 2212.5 \times 10^{-6} \] \[ = 2.2125 \times 10^{-3} \text{ radians} \] --- ### **Step 6: Convert Radians to Degrees** \[ \theta (\text{degrees}) = \theta (\text{radians}) \times \frac{180}{\pi} \] \[ = 2.2125 \times 10^{-3} \times 57.2958 \] \[ = 0.1267^\circ \] --- ## **Final Answer** ### **The Faraday rotation angle is approximately** \( \boxed{0.13^\circ} \). --- ### **Summary of Steps** 1. Wrote down the Faraday rotation formula for a plasma. 2. Plugged in all known constants and values. 3. Calculated numerator, denominator, and combined. 4. Converted the final answer to degrees. **Let me know if you need further clarification or the answer in a different form!**

The concept used in answering this question is the Faraday rotation effect in a plasma, which causes the plane of polarization of light passing through a magnetized plasma to rotate. The rotation angle depends on the strength of the magnetic field, the electron density, the wavelength of the probing light, and the path length through the plasma. By applying the known formula for Faraday rotation—derived from electromagnetic theory and plasma physics—and substituting the given parameters along with fundamental constants, we can calculate the expected rotation angle in degrees.