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

Step-by-step 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=(\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=(\pi d^(2))/(4)`

Substitute
`(\pi d^(2))/(4)[tex] for [tex]A` in equation (1).

`L=(\left(\mu_(0) N^(2)\right) (d d^(2))/(4))/(l)`

Substitute 580 for
`N, 4 \pi * 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}  </p><p>=\frac{4 \pi * 10^(-7) * 580 * 580 * \frac{x(8.00 \mathrm{~cm} * 8.00 \mathrm{~cm})}{4}}{36 \mathrm{~cm}} \\  </p><p>=5.90 \mathrm{mH}  </p><p>\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,

`</p><p>\varepsilon=L (d i)/(d t) </p><p>`

Here,

`\varepsilon` is the generated emf.

`L[tex] is the inductance of the coil. </p><p>[tex](d i)/(d t)[tex] is the rate of change of current. </p><p>Substitute [tex]5.90 \mathrm{mH}[tex] for [tex]L[tex] and [tex]4.00 \mathrm{~A} / \mathrm{s}[tex] for [tex](d i)/(d t)`

`\begin{array}{c} </p><p>\varepsilon=5.90 \mathrm{mH} * 4.00 \mathrm{~A} / \mathrm{s} \\ </p><p>= & 23.6 \mathrm{mV} </p><p>\end{array}`

Conclusion:

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