Question

The heating coils in a hair dryer are \(0.800{\rm{ }}cm\) in diameter, have a combined length of \(1.00{\rm{ }}m\), and a total of \(400\) turns. (a) What is their total self-inductance assuming they act like a single solenoid? (b) How much energy is stored in them when \(6.00{\rm{ }}A\) flows? (c) What average emf opposes shutting them off if this is done in \(5.00{\rm{ }}ms\) (one-fourth of a cycle for \(50{\rm{ }}Hz\) AC)?

Short answer

a. Their total self-inductance is\(31.25{\rm{ }}nH\).

b. The stored energy is\(562.5\;nJ\).

c. The induced emf is \(37.5\mu V\).

Step-by-step solution

Concept Introduction

A passive electrical component that dampens current variations is an inductor. Other names for inductors are coils and chokes. In electrical nomenclature, the letter\(L\)stands in for an inductor.

Information Given

• Diameter of wire coil:\(0.800{\rm{ }}cm\)
• Length of wire:\(1.00{\rm{ }}m\)
• Number of turns in coil:\(400\)
• The current value:\(6.00{\rm{ }}A\)
• The time value:\(5.00{\rm{ }}ms\)
• The frequency value: \(50{\rm{ }}Hz\)

Calculating Self-Inductance

a)

We need to be clear that we have a total length of\(S = 1\;m\)of wire with a diameter of\(d = 8\;mm\)coiled in\(N = 400\)turns. The entire length of the produced solenoid,\(I\), and its diameter,\(D\), must be determined. They are going to be

The inductance will be given by

\(\begin{align}{}S &= \frac{{{\mu _0}{N^2}\pi {D^2}}}{{4I}}\\ &= \frac{{{\mu _0}{N^2}\pi {S^2}}}{{4Nd{\pi ^2}{N^2}}}\\ &= \frac{{{\mu _0}{S^2}}}{{4\pi Nd}}\end{align}\)

Numerically, we will have

\(\begin{align}{}L &= \frac{{4\pi \times {{10}^{ - 7}} \times {1^2}}}{{4\pi \times 400 \times 0.008}}\\ &= 3.125 \times {10^{ - 8}}{\rm{ }}H\\ &= 31.25{\rm{ }}nH\end{align}\)

Therefore, the required solution is \(31.25{\rm{ }}nH\).

Calculating Stored Energy

b)

The stored energy will be

\(\begin{align}{}E &= \frac{{L{I^2}}}{2}\\ &= \frac{{3.125 \times {{10}^{ - 8}} \times {6^2}}}{2}\\ &= 5.625 \times {10^{ - 7}}\;J\end{align}\)

Therefore, the required solution is \(562.5\;nJ\)

Calculating Average EMF

c)

The induced emf will be given by

\(\begin{align}{}\varepsilon & = \frac{{L\Delta I}}{{\Delta t}}\\ &= \frac{{3.125 \times {{10}^{ - 8}} \times 6}}{{0.005}}\\ &= 3.75 \times {10^{ - 5}}\;V\end{align}\)

Therefore, the required solution is \(37.5\mu V\).