Consider two long co-axial solenoids $ S_1 $ and $ S_2 $, each of length $ l $ ($ l \gg r_2 $) and radius $ r_1 $ and $ r_2 $ ($ r_2r_1 $). The number of turns per unit length are $ n_1 $ and $ n_2 $, respectively. Derive an expression for mutual inductance $ M_{12} $ of solenoid $ S_1 $ with respect to solenoid $ S_2 $. Show that $ M_{21} = M_{12} $.

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Consider two long co-axial solenoids $ S_1 $ and $ S_2 $, each of length $ l $ ($ l \gg r_2 $) and radius $ r_1 $ and $ r_2 $ ($ r_2>r_1 $). The number of turns per unit length are $ n_1 $ and $ n_2 $, respectively. Derive an expression for mutual inductance $ M_{12} $ of solenoid $ S_1 $ with respect to solenoid $ S_2 $. Show that $ M_{21} = M_{12} $.

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Mutual Inductance of Two Coaxial Solenoids   Consider two long coaxial solenoids $ S_1 $ and $ S_2 $, each of length $ l $ ($ l \gg r_2 $) and radius $ r_1 $ and $ r_2 $ ($ r_2 > r_1 $), respectively. The number of turns per unit length are $ n_1 $ and $ n_2 $, respectively. We will derive the expression for the mutual inductance $ M_{12} $ of solenoid $ S_1 $ with respect to solenoid $ S_2 $ and show that $ M_{21} = M_{12} $. Step 1: Magnetic Field Produced by Solenoid $ S_1 $ The magnetic field inside a long solenoid is given by: $$ B_1 = \mu_0 n_1 I_1 $$ where: - $ \mu_0 $ is the permeability of free space, - $ n_1 $ is the number of turns per unit length of solenoid $ S_1 $, - $ I_1 $ is the current flowing through solenoid $ S_1 $. This magnetic field is directed along the axis of the solenoid and is uniform inside the solenoid. Step 2: Magnetic Flux through Solenoid $ S_2 $ The magnetic flux $ \Phi_2 $ through the area of one turn of solenoid $ S_2 $ is given by: $$ \Phi_2 = B_1 A_2 $$ where: - $ A_2 = \pi r_2^2 $ is the cross-sectional area of solenoid $ S_2 $, - $ B_1 = \mu_0 n_1 I_1 $ is the magnetic field produced by solenoid $ S_1 $. Therefore, the total flux through $ N_2 $ turns of solenoid $ S_2 $ is given by: $$ \Phi_2 = \mu_0 n_1 I_1 \pi r_2^2 l $$ where $ l $ is the length of solenoid $ S_2 $. Step 3: Induced EMF in Solenoid $ S_2 $ The induced EMF $ \mathcal{E}_2 $ in solenoid $ S_2 $ due to the time-varying magnetic flux is given by Faraday’s law: $$ \mathcal{E}_2 = - \frac{d\Phi_2}{dt} = - \frac{d}{dt} \left( \mu_0 n_1 I_1 \pi r_2^2 l \right) $$ Since $ I_1 $ is the current in solenoid $ S_1 $, we can write: $$ \mathcal{E}_2 = - M_{12} \frac{dI_1}{dt} $$ where $ M_{12} $ is the mutual inductance between solenoid $ S_1 $ and solenoid $ S_2 $. Thus, comparing both expressions for the induced EMF, we get: $$ M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l $$ This is the expression for the mutual inductance between the two solenoids. Step 4: Symmetry of Mutual Inductance From the above derivation, it is evident that the mutual inductance is symmetric. That is, the mutual inductance $ M_{12} $ of solenoid $ S_1 $ with respect to solenoid $ S_2 $ is equal to the mutual inductance $ M_{21} $ of solenoid $ S_2 $ with respect to solenoid $ S_1 $. Therefore, we can write: $$ M_{21} = M_{12} $$ Conclusion: The mutual inductance between two coaxial solenoids is given by: $$ M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l $$ and it is symmetric, i.e., $ M_{21} = M_{12} $.

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Solution and Explanation

Mutual Inductance of Two Coaxial Solenoids

Step 1: Magnetic Field Produced by Solenoid \( S_1 \)

Step 2: Magnetic Flux through Solenoid \( S_2 \)

Step 3: Induced EMF in Solenoid \( S_2 \)

Step 4: Symmetry of Mutual Inductance

Conclusion:

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