Question

A $\mathbf{2}\mathbf{.}\mathbf{50}\mathit{m}\mathit{H}$toroidal solenoid has an average radius of $\mathbf{6}\mathbf{.}\mathbf{00}\mathit{c}\mathit{m}$and a cross-sectional area of $\mathbf{2}\mathbf{.}\mathbf{00}\mathit{V}$. (a) How many coils does it have? (Make the same assumption as in Example 30.3.) (b) At what rate must the current through it change so that a potential difference of $\mathbf{2}\mathbf{.}\mathbf{00}\mathit{V}$is developed across its ends?

Short answer

(a) the toroidal solenoid has$1940$coils.

(b) the current rate for potential difference to be$2.00V$developed across its ends is$800A/s$

Step-by-step solution

Define self-induction and self-inductance of toroidal solenoid

Induction of an electromotive force in a circuit by a varying current in the same circuit, is called self-induction.

Self-inductance is measure of self-induction in circuit. It is expressed as,

$\epsilon =-L\frac{dI}{dt}$where $\epsilon$is induced emf, $\frac{dI}{dt}$is rate at which current changes and L is the self-inductance.

Self-inductance of toroidal solenoidal is,

$\mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{N}}^{\mathbf{2}}\mathbf{A}}{\mathbf{2}\mathbf{\pi }\mathbf{r}}$where $L$is self-inductance of toroid having N turns, $A$area of cross-section and $2\mathrm{\pi r}$be circumference.

Apply the formulae

(a) The self-inductance $L=2.50mH$toroidal solenoid has an average radius of $r=6.00cm$and a cross-sectional area of $A=2.00c{m}^2$.

The number of coils $N$in the toroid is,

$$\begin{gathered} L=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2\mathrm{A}}{2\mathrm{\pi r}} \\ N=\sqrt{\frac{2\mathrm{\pi }\times 0.0600\times 2.50\times {10}^{-3}}{4\mathrm{\pi }\times {10}^{-7}\times 2.00\times {10}^{-4}}} \\ N=1940 \end{gathered}$$

(b) The magnitude of potential difference across the coil is expressed as,

$\left|\epsilon \right|=L\frac{dI}{dt}$

$\frac{dI}{dt}=\frac{\epsilon }{L}$

$$\begin{gathered} \frac{dI}{dt}=\frac{2.00}{2.50\times {10}^{-3}} \\ \frac{dI}{dt}=800A/s \end{gathered}$$

From above, current rate required to have $\epsilon =2.00V$potential difference is given as,

Therefore, (a) the toroidal solenoid has$1940$coils;(b) the current rate for potential difference to be$2.00V$developed across its ends is$800A/s$.