Question

The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by wrapping this wire in a spiral around a 2.3-mm-diameter glass core.How many turns of wire are needed?

Answer 1

Explanation:

The given data is as follows.

         Resistance (R) = 1200 ohm,     Area (A) = $20 \times 10^{-6}$ m (as $1 \mu m = 10^{-6} m$)

            Diameter (d) = 2.3 mm = $2.3 \times 10^{-3}$ m

First, we will calculate the length as follows.

            R = $\rho \frac{L}{A}$

Here,  $\rho$ = resistivity of aluminium = $2.65 \times 10^{-8}$

Putting the given values above and we will calculate the value of length as follows.

               R = $\rho \frac{L}{A}$

             1200 = $2.65 \times 10^{-8} \times \frac{L}{20 \times 10^{-6}}$

               L = $9.056 \times 10^{5}$

As the circumference of circular wire = $2 \pi r$

or,                                                          = $2 \times \pi \times \frac{d}{2}$  

                                                              = $\pi \times d$

And, number of turns will be calculated as follows.

             No. of turns × Circumference = Length of wire

              No. of turns × $3.14 \times 2.3 \times 10^{-3} = 9.056 \times 10^{5}$

                               = $1.25 \times 10^{8}$

Thus, we can conclude that $1.25 \times 10^{8}$  turns of wire are needed.