Explanation:
Given info: The Number of turns in the wire is 580 the diameter of
tube is [tex]8.00 \mathrm{~cm}[/tex] and the length of the tube up to which wire is wrapped is [tex]36.0 \mathrm{~cm}[/tex].
Formula to calculate the inductance of the coil is,
[tex]L=\frac{\mu_{0} N^{2} A}{l}[/tex]
Here,
[tex]L[/tex] is inductance of the coil.
[tex]\mu_{0}[/tex] is the permittivity.
[tex]N[/tex] is the number of turns.
[tex]l[/tex] is the length up to which wire is wrapped.
[tex]A[/tex] is the cross sectional area of the coil.
The expression for the area is,
[tex]A=\frac{\pi d^{2}}{4}[/tex]
Substitute [tex]\frac{\pi d^{2}}{4}[tex] for [tex]A[/tex] in equation (1).
[tex]L=\frac{\left(\mu_{0} N^{2}\right) \frac{d d^{2}}{4}}{l}[/tex]
Substitute 580 for [tex]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 .[/tex]
[tex]\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} [/tex]
Conclusion:
Therefore, the inductance of the given single conductor wire is
[tex]5.90 \mathrm{mH} .[/tex]
Given info: The rate of increasing current [tex]4.00 \mathrm{~A} / \mathrm{s}[/tex] and the inductance of the coil [tex]5.90 \mathrm{mH}[/tex].
The generated emf is,
[tex]
\varepsilon=L \frac{d i}{d t}
[/tex]
Here,
[tex]\varepsilon[/tex] is the generated emf.
[tex]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}[/tex]
[tex]\begin{array}{c}
\varepsilon=5.90 \mathrm{mH} \times 4.00 \mathrm{~A} / \mathrm{s} \\
= & 23.6 \mathrm{mV}
\end{array}[/tex]
Conclusion:
Therefore, the generated emf is [tex]23.6 \mathrm{mV}[/tex].
