Question

You are working as an electrical technician. One day, out in CR the field, you need an inductor but cannot find one. Look- ing in your wire supply cabinet, you find a cardboard tube with single-conductor wire wrapped uniformly around it to form a solenoid. You carefully count the turns of wire and find that there are 580 turns. The diameter of the tube is 8.00 cm, and the length of the wire-wrapped portion is 36.0 cm. You pull out your calculator to determine:
a. the inductance of the coil
b. the emf generated in it if the current in the wire increases at the rate of 4.00 A/s.

Answer 1

Explanation:

Given info: The Number of turns in the wire is 580 the diameter of  

tube is $8.00 \mathrm{~cm}$ and the length of the tube up to which wire is wrapped is $36.0 \mathrm{~cm}$.  

Formula to calculate the inductance of the coil is,  

$L=\frac{\mu_{0} N^{2} A}{l}$  

Here,  

$L$ is inductance of the coil.  

$\mu_{0}$ is the permittivity.  

$N$ is the number of turns.  

$l$ is the length up to which wire is wrapped.  

$A$ is the cross sectional area of the coil.  

The expression for the area is,  

$A=\frac{\pi d^{2}}{4}$

Substitute $\frac{\pi d^{2}}{4}[tex] for [tex]A$ in equation (1).  

$L=\frac{\left(\mu_{0} N^{2}\right) \frac{d d^{2}}{4}}{l}$

Substitute 580 for $N, 4 \pi \times 10^{-7}[tex] for [tex]\mu_{0}, 8.00 \mathrm{~cm}[tex] for [tex]d[tex] and [tex]36.0 \mathrm{~cm}[tex] for [tex]l .$  

$\begin{array}{c} =\frac{4 \pi \times 10^{-7} \times 580 \times 580 \times \frac{x(8.00 \mathrm{~cm} \times 8.00 \mathrm{~cm})}{4}}{36 \mathrm{~cm}} \\ =5.90 \mathrm{mH} \end{array}$

Conclusion:  

Therefore, the inductance of the given single conductor wire is  

$5.90 \mathrm{mH} .$

Given info: The rate of increasing current $4.00 \mathrm{~A} / \mathrm{s}$ and the inductance of the coil $5.90 \mathrm{mH}$.

The generated emf is,

$\varepsilon=L \frac{d i}{d t}$

Here,

$\varepsilon$ is the generated emf.

$L[tex] is the inductance of the coil. [tex]\frac{d i}{d t}[tex] is the rate of change of current. Substitute [tex]5.90 \mathrm{mH}[tex] for [tex]L[tex] and [tex]4.00 \mathrm{~A} / \mathrm{s}[tex] for [tex]\frac{d i}{d t}$

$\begin{array}{c} \varepsilon=5.90 \mathrm{mH} \times 4.00 \mathrm{~A} / \mathrm{s} \\ = & 23.6 \mathrm{mV} \end{array}$

Conclusion:

Therefore, the generated emf is $23.6 \mathrm{mV}$.