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Q62PE

Expert-verifiedFound in: Page 862

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) Calculate the self-inductance of a \({\rm{50}}{\rm{.0}}\)cm long, \({\rm{10}}{\rm{.0}}\)cm diameter solenoid having \({\rm{1000}}\) loops. (b) How much energy is stored in this inductor when \({\rm{20}}{\rm{.0}}\) A of current flows through it? (c) How fast can it be turned off if the induced emf cannot exceed \({\rm{3}}{\rm{.00}}\)V?**

(a.) The inductor's self-inductance is \({\rm{19}}{\rm{.7mH}}\).

(b.) The inductor has \({\rm{3}}{\rm{.95\;J}}\) of energy stored in it.

(c.) The shortest time in which the emf will be less than \(3\;{\rm{V}}\) is \({\rm{131\;ms}}{\rm{.}}\)

- Length of the coil: \(l = 50.0\;{\rm{cm}}\).
- Diameter of the coil: \(d = 10.0\;{\rm{cm}}\).
- Number of loops: \(N = 1000\)
- Current flowing through the inductor coil: \(\Delta I = 20.0 {\rm{A}}\).
- Maximum Induced emf: \(\varepsilon = \;3.0\;{\rm{V}}\)

**If a solenoid is connected to an A.C. source, due to the continuously changing magnetic field produced by the solenoid, an emf is induced in the same solenoid. This induced emf is such that it opposes the change in the magnetic field. This phenomenon is called self-inductance.**

a)

The inductor's self-inductance is calculated as

\(L = \frac{{{\mu _0}{N^2}A}}{l}\)

where \(A = \frac{{\pi {D^2}}}{4}\)is the cross-section area, \(N\) is the number of turns, and \(l\) is the coil length. As a result, we will have-

\(\begin{aligned} L &= \frac{{\pi {\mu _0}{N^2}{D^2}}}{{4l}}\\ &= \frac{{\pi \times \left( {4\pi \times {{10}^7}\;{\rm{H}}{{\rm{m}}^{{\rm{ - 1}}}}} \right) \times {{1000}^2} \times {{\left( {0.1\;{\rm{m}}} \right)}^2}}}{{4 \times \left( {0.5\;{\rm{m}}\;} \right)}}\\ &= 1.97 \times {10^{ - 2}}\;{\rm{H}}\end{aligned}\)

Therefore, the self-inductance of the solenoid is \(1.97 \times {10^{ - 2}}\;{\rm{H}}\)

b)

When the inductance and current are known, the energy stored in an inductor is given by

\(E = \frac{{L{I^2}}}{2}\)

Numerically in our case we will have

\(\begin{aligned} {\rm{E }}&=\frac{{\left( {{\rm{0}}{\rm{.0197}}\;{\rm{H}}} \right){\rm{ \times }}{{\left( {{\rm{20}}\;{\rm{A}}} \right)}^{\rm{2}}}}}{{\rm{2}}}\\{\rm{ }}&={\rm{.94\;J}}\end{aligned}\)

Therefore, energy stored in the inductor is \({\rm{3}}{\rm{.94\;J}}\)

c)

if we ignore the sign, the induced electromotive force is given as-

\(\varepsilon = \frac{{L\Delta I}}{{\Delta t}}\)

the shortest time\(\Delta t\) for the emf not to surpass a given amount \(\varepsilon \) is-

\(\Delta t = \frac{{L\Delta I}}{\varepsilon }.\)

Numerically, we will have

\(\begin{aligned} \Delta t &= \frac{{0.0197\;{\rm{H}} \times 20\;{\rm{A}}}}{{3\;{\rm{V}}}}\\ &= 0.13\;{\rm{s}}\end{aligned}\)

Therefore, the shortest time in which the emf will not exceed \(3\;{\rm{V}}\) is \({\rm{0}}{\rm{.131\;s}}\).

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