91Ó°ÊÓ

Faraday’s law states that a current is induced in a conductor when it is exposed to a time varying magnetic flux. This induced current is driven by a force called electromotive or electromagnetic force. The magnitude of induced emf is given by

$\mathit{Î\mu }\mathbf{=}\mathbf{-}\mathit{L}\frac{\mathbf{}\mathbf{di}}{\mathbf{dt}}$

Where L is the inductance of the conductor.

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it.

Lenz further explained the direction of this induced current. According to lenz, the direction of induced current will be such that the magnetic field created by the induced current opposes the changing magnetic field which caused its induction.

We are given,

The inductance of solenoid, L = 0.100 mH

Cross sectional area, A = 0.600 cm2

Mean radius of solenoid, r = 12.0 cm

Self-Inductance of a toroidal solenoid is given by

$$\begin{gathered} L=\frac{{\mathrm{μ}}_0{N}^{2}A}{2\mathrm{Ï€°ù}} \\ N=\sqrt{\frac{2\mathrm{Ï€°ù}L}{{\mathrm{μ}}_{0}A}} \\ =\sqrt{\frac{\left(0.120m\right)\left({10}^{-4}H\right)}{\left(2*{10}^{-7}Tm/A\right)\left(0.600*{10}^{-4}{m}^2\right)}} \\ =1000turns \end{gathered}$$

Therefore, the number of turns in the toroidal solenoid is 1000 turns.

The resistance per unit length of the wire is given so if the length of the wire is known then resistance of the wire can be determined easily.

Let L be the length of the wire, and c be the circumference of one turn then,

$$\begin{gathered} L=N.c \\ =N\left(2\mathrm{Ï€°ù}\right) \\ =2\mathrm{\pm ·Ï€}\sqrt{\frac{\mathrm{A}}{\mathrm{Ï€}}} \\ =2\mathrm{N}\sqrt{\mathrm{Ï€´¡}} \\ =2*1000*\sqrt{\mathrm{Ï€}0.600*{10}^{-4}{\mathrm{m}}^2} \\ =27.46\mathrm{m} \end{gathered}$$

Thus, the resistance of the wire = (0.0760 Ω) x (27.46 m) = 2.09 Ω

Therefore, the resistance of the toroidal solenoid is 2.09 Ω.