Question

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

Short answer

Answer: \(B_2 = \mu_0 n \left(\frac{N \left(\mu_0 n i_0 A_2\right)(-\omega \sin \omega t)}{R}\right)\)

Step-by-step solution

Calculate the magnetic field produced by the outer solenoid

The magnetic field inside a solenoid can be calculated using the formula: \(B = \mu_0 n i\) Where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space \((4 \pi \times 10^{-7} Tm/A)\), \(n\) is the number of turns per unit length, and \(i\) is the current flowing through the solenoid. In this case, the current in the outer solenoid is given by \(i = i_0 \cos \omega t\). Therefore, the magnetic field inside the outer solenoid is: \(B_1 = \mu_0 n i_0 \cos \omega t\)

Calculate the induced emf in the inner solenoid

According to Faraday's law of electromagnetic induction, the induced emf in the inner solenoid can be calculated using the formula: \(\epsilon = - N \frac{d\Phi}{dt}\) Where \(\epsilon\) is the induced emf, \(N\) is the number of turns in the inner solenoid, and \(\Phi\) is the magnetic flux through the inner solenoid. The magnetic flux can be calculated using the formula: \(\Phi = B A\) Where \(B\) is the magnetic field and \(A\) is the cross-sectional area. In this case, since both solenoids have the same number of turns, the magnetic field in the inner solenoid due to the outer solenoid is the same as the magnetic field inside the outer solenoid, which we calculated in Step 1. Therefore, the magnetic flux through the inner solenoid is: \(\Phi = B_1 A_2 = \mu_0 n i_0 A_2 \cos \omega t\) Now, we can differentiate the magnetic flux with respect to time and substitute that into Faraday's law to obtain the induced emf in the inner solenoid: \(\frac{d\Phi}{dt} = \mu_0 n i_0 A_2 \left(-\omega \sin \omega t\right)\) \(\epsilon = - N \left(\mu_0 n i_0 A_2\right) \left(-\omega \sin \omega t\right)\)

Calculate the induced current in the inner solenoid

The induced current in the inner solenoid can be calculated using Ohm's law: \(I = \frac{\epsilon}{R}\) Substituting the induced emf from Step 2, we get the induced current in the inner solenoid: \(I_2 = \frac{N \left(\mu_0 n i_0 A_2\right)(-\omega \sin \omega t)}{R}\)

Calculate the magnetic field inside the inner solenoid due to the induced current

Using the formula for the magnetic field inside a solenoid from Step 1, we can calculate the magnetic field inside the inner solenoid due to the induced current: \(B_2 = \mu_0 n I_2\) Substituting the induced current from Step 3, we get the magnetic field inside the inner solenoid: \(B_2 = \mu_0 n \left(\frac{N \left(\mu_0 n i_0 A_2\right)(-\omega \sin \omega t)}{R}\right)\) This is the expression for the magnetic field in the inner solenoid due to the induced current.