Question

An ideal solenoid ( \(N_{1}\) turns, length \(L_{1},\) radius \(r_{1}\) ) is placed inside another ideal solenoid ( \(N_{2}\) turns, length \(L_{2}>L_{1}\), radius \(r_{2}>r_{1}\) ) such that the axes of the two coincide. (a) What is the mutual inductance? (b) If the current in the outer solenoid is changing at a rate \(\Delta I_{2} / \Delta t,\) what is the magnitude of the induced emf in the inner solenoid?

Short answer

Question: Calculate the magnitude of the induced emf in the inner solenoid if the mutual inductance between the solenoids is 1.5 mH and the rate of change of current in the outer solenoid is 3.0 A/s. Answer: To find the magnitude of the induced emf in the inner solenoid, we use the formula: \(|\text{emf}| = M |\frac{dI_2}{dt}|\) Where M is the mutual inductance and \(\frac{dI_2}{dt}\) is the rate of change of current in the outer solenoid. Given values: M = 1.5 mH = 1.5 x 10^(-3) H \(\frac{dI_2}{dt}\) = 3.0 A/s Substitute the given values into the formula: \(|\text{emf}| = (1.5 × 10^{-3}) × (3.0)\) \(|\text{emf}| = 4.5 × 10^{-3}\) The magnitude of the induced emf in the inner solenoid is 4.5 mV.

Step-by-step solution

(a) Calculating the mutual inductance

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: \(M = \frac{\Phi_{1,\text{total}}}{I_2} = \mu_{0} \frac{N_1 N_2}{L_2} \pi r_1^2\)

(b) Induced emf in the inner solenoid

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.