Physics, 19.03.2021 18:30 stupidsmoke4272
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.
Answers: 2
and the length of the tube up to which wire is wrapped is 
. Â 
 
is inductance of the coil. Â 
is the permittivity. Â 
is the number of turns. Â 
is the length up to which wire is wrapped. Â 
is the cross sectional area of the coil. Â 
![\frac{\pi d^{2}}{4}[tex] for [tex]A](/tpl/images/1206/9843/1a871.png)
in equation (1). Â 
![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 .](/tpl/images/1206/9843/11e85.png)
 
and the inductance of the coil 
.
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}](/tpl/images/1206/9843/ce1d1.png)
.
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