A toroidal solenoid has 2900 closely wound turns, cross-sectional area 0.450 cm², mean radius 9.00 cm, and resistance R = 2.80 Ω. Ignore the variation of the magnetic field across the cross section of the solenoid. What is the amplitude of the current in the solenoid if it is connected to an ac source that has voltage amplitude 24.0 V and frequency 495 Hz?

A toroid is a coil of insulated wire wound around in a doughnut-shaped form.The inductance due to a toroidal solenoid is given as$L=\frac{{\mu }_{0}A{N}^2}{2\mathrm{\pi r}}$ where, L is the${\mu }_0$ inductance, is the permeability of free space,A is the area of cross section of the toroid,N is the number of turns in the solenoid,r is the mean radius of the toroid.

In the question it is mentioned that the solenoid has a resistance, which is quite practical. But,it can be considered that the resistance is in series with a pure inductor.Thus the impedance of thee circuit comes out to be $Z=\sqrt{R^2+X_L^2}$, where, is the resistance of the solenoid, and$X_L$ is the inductive reactance given by, $X_L=\omega L$, where,$\omega$ is the angular frequency of the A/C signal.

Given:

$$\begin{gathered} {\mu }_0=4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m} \\ \mathrm{A}=0.45\times {10}^{-4}{\mathrm{m}}^2 \\ \mathrm{N}=2900 \\ \mathrm{r}=9\times {10}^{-2}\mathrm{m} \end{gathered}$$

Hence, the value of the inductance is-

$$\begin{gathered} \mathrm{L}=\frac{4\mathrm{\pi }\times -{10}^{-7}\times 0.45\times {10}^{-4}\times \left(2900\right)}{2\mathrm{\pi }\times 9\times {10}^{-2}} \\ =8.41\times {10}^{-4}\mathrm{H} \end{gathered}$$

Hence the inductance of the toroidal solenoid is$8.41\times {10}^{-4}\mathrm{H}$

Given:

$$\begin{gathered} \mathrm{R}=2.8\mathrm{\Omega } \\ \mathrm{\omega }=2\mathrm{\pi }\times 495 \\ =3110.18\mathrm{rad}/\mathrm{s} \end{gathered}$$

The reactance of the inductor can be calculated as-

$$\begin{gathered} X_L=\omega L \\ =3110.18\times 8.41\times {10}^{-4} \\ =2.616\Omega \end{gathered}$$

Hence, the impedance of the circuit is

$$\begin{gathered} \mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2} \\ =\sqrt{2.8^2+2.{616}^2} \\ =3.83\mathrm{\Omega } \end{gathered}$$

Hence, the impedance of the circuit is$3.83\Omega$

The amplitude of the current refers to the peak value of current which can be calculated as $I_{peak}=\frac{V_{peak}}{Z}$, where$I_{peak}$ is the peak value of current,$v_{peak}$ is the peak value of voltage. Therefore,

$$\begin{gathered} I_{peak}=\frac{24}{3.83} \\ =6.27A \end{gathered}$$

Hence, the amplitude of current in the circuit is 6.27 A