91Ó°ÊÓ

To find the mutual inductance, M, between the two solenoids, we need to calculate the magnetic flux linkage in the inner solenoid due to the magnetic field produced by the outer solenoid. The magnetic field inside the outer solenoid can be calculated using its properties: \(B_2 = \mu_{0} \frac{N_2}{L_2} I_2\) Since the inner solenoid is entirely within the outer solenoid, the magnetic field inside it will be approximately uniform. The magnetic flux through each turn of the inner solenoid will be: \(\Phi_1 = B_2 \pi r_1^2\) The total magnetic flux linkage in the inner solenoid is: \(\Phi_{1,\text{total}} = N_1 \Phi_1 = N_1 B_2 \pi r_1^2\) Mutual inductance M is defined as the ratio of the magnetic flux linkage in the inner solenoid to the current in the outer solenoid: \(M = \frac{\Phi_{1,\text{total}}}{I_2} = \mu_{0} \frac{N_1 N_2}{L_2} \pi r_1^2\)

We can find the induced emf in the inner solenoid using Faraday's law of electromagnetic induction, which states that the induced emf in a loop is negative the rate of change of magnetic flux linkage through the loop: \(\text{emf} = -\frac{d\Phi_{1,\text{total}}}{dt}\) We can write the magnetic flux linkage as a function of the current I2 by replacing \(B_2\) with its expression in terms of \(I_2\): \(\Phi_{1,\text{total}} = N_1 \mu_{0} \frac{N_2}{L_2} \pi r_1^2 I_2\) Now we can find the derivative with respect to time: \(\frac{d\Phi_{1,\text{total}}}{dt} = N_1 \mu_{0} \frac{N_2}{L_2} \pi r_1^2 \frac{dI_2}{dt}\) The induced emf in the inner solenoid is: \(\text{emf} = - \frac{d\Phi_{1,\text{total}}}{dt} = N_1 \mu_{0} \frac{N_2}{L_2} \pi r_1^2 \frac{dI_2}{dt}\) We want the magnitude of the induced emf, which is given by: \(|\text{emf}| = N_1 \mu_{0} \frac{N_2}{L_2} \pi r_1^2 |\frac{dI_2}{dt}|\) \(|\text{emf}| = M |\frac{dI_2}{dt}|\) So the magnitude of the induced emf in the inner solenoid is equal to the mutual inductance M times the magnitude of the rate of change of the current in the outer solenoid.