Q15P - Fundamentals Of Physics

Question

A coil is formed by winding 250turns of insulated 16-gauge copper wire (diameter = 1.3 mm) in a single layer on a cylindrical form of radius 12cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Use Table 26-1.)

Step-by-Step Solution

Answer

The resistance of the coil is 2.4$\mathrm{\Omega }$.

Step 1: The given data

1. Number of turns of coil,n = 250
2. Diameter of the wire, $d=1.3 \mathrm{mm} \mathrm{or} 1.3\times {10}^{-3}\mathrm{m}$
3. Radius of coil, $R=12 \mathrm{cm} \mathrm{or} 12\times {10}^{-2}\mathrm{m}$

Step 2: Understanding the concept of the resistance

The voltage is directly proportional to the current and the constant of proportionality is called Resistance. This is called Ohm’s law.

We can find the total length of the coil by using the radius of the coil. We can also find the area of the cross-section of the wire from its diameter. Then, by using these values in the formula for resistance, we can find the resistance of the copper coil.

 

Formulae:

The resistance of a wire due, $R=\frac{pL}{A}$                                                                           …(i)

Here, R is resistance, p is resistivity, L is length, A is the area of cross-section.

The circumference of the circle, $L=2\mathrm{\pi }R$                                                                     …(ii) 

is circumference, is the radius of the circle.

The cross-sectional area of a circle, $A={\mathrm{\pi R}}^2$                                                             …(iii)

Step 3: Calculation of the resistance of the coil

Length of one turn of the coil can be given using the circumference of the circle.

So, the length of 250 turns of the coil using equation (ii) would be,

$$\begin{gathered} L=250\times \left(2\mathrm{\pi R}\right) \\ =250\times 2(3.142)(12\times {10}^{-2}\mathrm{m}) \\ =188.49\mathrm{m} \end{gathered}$$

Similarly, the cross sectional area of the wire is given using equation (iii) as follows:

$$\begin{gathered} A=\mathrm{\pi }\frac{{\mathrm{d}}^2}{4} (\therefore \mathrm{r}=\mathrm{d}/2) \\ =(3.142)\frac{{(1.3\times {10}^{-3}\mathrm{m})}^2}{4} \\ =1.327\times {10}^{-6}{\mathrm{m}}^2 \end{gathered}$$ 

Resistivity of copper is $\mathrm{p}=1.69\times {10}^{-8}\mathrm{\Omega m}$

Now, using these above values in equation (i); the resistance of the copper coil can be calculated as follows:

$$\begin{gathered} \mathrm{R}=\frac{(1.69\times {10}^{-8}\mathrm{\Omega m})(188.49\mathrm{m})}{1.327\times {10}^{-6}{\mathrm{m}}^2} \\ =2.4\mathrm{\Omega } \end{gathered}$$

Therefore, the resistance of the copper coil is $2.4\mathrm{\Omega }$.