Question

A long solenoid with cross-sectional area \(A_{1}\) surrounds another long solenoid with cross-sectional area \(A_{2}

Short answer

Answer: The expression for the magnetic field in the inner solenoid due to the induced current from the outer solenoid is \(B_2 = \frac{\mu_0^2 n^2 A_2 i_0 \omega \sin (\omega t)}{R}\), where \(B_2\) is the magnetic field in the inner solenoid, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, \(A_2\) is the cross-sectional area of the inner solenoid, \(i_0\) is the amplitude of the current in the outer solenoid, \(\omega\) is the angular frequency, \(t\) is time, and \(R\) is the resistance of the inner solenoid.

Step-by-step solution

Find the magnetic field produced by the outer solenoid

The current in the outer solenoid is given by \(i=i_0 \cos{\omega t}\). Using Ampere's Law, the magnetic field produced by the outer solenoid can be calculated as follows: \(B_1 = \mu_0 n i_0 \cos (\omega t)\) Here, \(B_1\) represents the magnetic field produced by the outer solenoid, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(i_0\) is the amplitude of the current.

Find the induced EMF in the inner solenoid

To find the induced EMF in the inner solenoid, we will use Faraday's Law, which states that the induced EMF is equal to the negative rate of change of the magnetic flux through the solenoid: \(EMF = -\frac{d\Phi}{dt}\) Since the inner solenoid is surrounded by the outer solenoid, the magnetic flux through the inner solenoid is equal to the product of the magnetic field produced by the outer solenoid and the cross-sectional area of the inner solenoid: \(\Phi = B_1 A_2\) Now, taking the derivative with respect to time: \(EMF = -\frac{d}{dt} (B_1 A_2) = - A_2 \frac{d}{dt} (\mu_0 n i_0 \cos (\omega t))\) \(EMF = A_2 \mu_0 n i_0 \omega \sin (\omega t)\)

Find the induced current in the inner solenoid

The induced current in the inner solenoid is related to the induced EMF and the resistance of the inner solenoid using Ohm's Law: \(I_2 = \frac{EMF}{R}\) Substituting the previously calculated EMF value: \(I_2 = \frac{A_2 \mu_0 n i_0 \omega \sin (\omega t)}{R}\)

Find the magnetic field in the inner solenoid due to the induced current

Now that we have found the induced current in the inner solenoid, we can find the magnetic field in the inner solenoid due to the induced current using Ampere's Law: \(B_2 = \mu_0 n I_2\) Substituting the value for \(I_2\): \(B_2 = \mu_0 n \frac{A_2 \mu_0 n i_0 \omega \sin (\omega t)}{R}\) Simplifying, we get the final expression for the magnetic field in the inner solenoid due to the induced current: \(B_2 = \frac{\mu_0^2 n^2 A_2 i_0 \omega \sin (\omega t)}{R}\)