Two solenoids, of \(N_{1}\) and \(N_{2}\) turns respectively, are wound on the same form. They have the same length \(L\) and radius \(r\) (a) What is the mutual inductance of these two solenoids? (b) If an ac current $$I_{1}(t)=I_{\mathrm{m}} \sin \omega t$$ flows in solenoid $$1\left(N_{1} \text { turns }\right)$$ write an expression for the total flux through solenoid \(2 .\) (c) What is the maximum induced emf in solenoid \(2 ?[\text { Hint: Refer to Eq. }(20-7) .]\)

The mutual inductance M can be calculated using the formula: $$M = \frac{\mu_0 N_1 N_2 A}{L}$$ where $$\mu_0$$ is the permeability of free space, $$N_1$$ and $$N_2$$ are the number of turns of the two solenoids, A is the cross-sectional area of the solenoids, and L is their length. The cross-sectional area can be calculated using the radius r: $$A = \pi r^2$$ Substituting this into the formula for M, we get: $$M = \frac{\mu_0 N_1 N_2 \pi r^2}{L}$$ This gives us the mutual inductance for the two solenoids.

The magnetic field inside solenoid 1 is given by: $$B_1 = \mu_0 n_1 I_1(t)$$ where $$n_1 = \frac{N_1}{L}$$ is the turns density of solenoid 1, and $$I_1(t) = I_m \sin \omega t$$ is the current through solenoid 1. The flux through each turn in solenoid 2 due to the magnetic field in solenoid 1 is the product of the magnetic field B1 and the cross-sectional area A: $$\Phi_{2 \text{ per turn}} = B_1 A = \mu_0 n_1 I_1(t) A$$ The total flux through solenoid 2 is the sum of the flux through each turn, so $$\Phi_2 = N_2 \Phi_{2 \text{ per turn}} = N_2 \mu_0 n_1 I_1(t) A$$ Substitute the given current and area expressions, to get: $$\Phi_2 = N_2 \mu_0 n_1 I_m \sin \omega t \pi r^2$$ This gives us the total flux through solenoid 2 as a function of time.

The induced emf in solenoid 2 can be calculated using Faraday's law: $$e_2 = - N_2 \frac{d \Phi_2}{d t}$$ Differentiate the expression for $$\Phi_2$$ with respect to time: $$\frac{d\Phi_2}{dt} = N_2 \mu_0 n_1 I_m \omega \cos \omega t \pi r^2$$ Substituting this into the formula for the induced emf in solenoid 2: $$e_2 = - N_2 \left( N_2 \mu_0 n_1 I_m \omega \cos \omega t \pi r^2 \right)$$ The maximum induced emf occurs when $$\cos \omega t = 1$$, so $$e_{2 \text{ max}} = N_2 \left( N_2 \mu_0 n_1 I_m \omega \pi r^2 \right)$$ This gives us the maximum induced electromotive force in solenoid 2.