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College Physics (Urone)
Found in: Page 862
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

The \(4.00{\rm{ }}A\) current through a \(7.50{\rm{ }}mH\) inductor is switched off in \(8.33{\rm{ }}ms\). What is the emf induced opposing this?

The \(4.00{\rm{ }}A\) current through a \(7.50{\rm{ }}mH\) inductor is switched off in \(8.33{\rm{ }}ms\). The emf opposing this is, \(\varepsilon = - 3.6\;V\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Concept Introduction

A current produced by the creation of voltage (electromotive force) in the presence of a shifting magnetic field is known as electromagnetic induction.

This happens when a conductor is placed in a magnetic field that is moving (using an AC power source) or when a conductor is continuously moving in a magnetic field that is stationary.

Step 2: Information Provided

  • Inductance of the inductor:

\(\begin{array}{c}M = 7.50{\rm{ }}mH = \frac{{7.50}}{{1000}}\\ = 0.0075{\rm{ }}H\end{array}\).

  • Current through an inductor: \(4.00{\rm{ }}A\).
  • Time in which current is switched off:

\(\begin{array}{c}t = 8.33{\rm{ }}ms = \frac{{8.33}}{{1000}}\\ = 0.00833{\rm{ }}s\end{array}\).

Step 3: Calculation for the voltage

The formula for the induced electromotive force is –

\(\varepsilon = - M\frac{{\Delta I}}{{\Delta t}}\)

After substituting the values, the calculation is –

\(\begin{array}{c}\varepsilon = - 0.0075 \cdot \frac{4}{{0.00833}}\\ = - 3.6\;V\end{array}\)

Therefore, the value for emf is obtained as \(\varepsilon = - 3.6\;V\).

Most popular questions for Physics Textbooks

Approximately how does the emf induced in the loop in Figure 23.57 (b) depend on the distance of the center of the loop from the wire?

Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of in one is switched off in , inducing an emf in the other. What is their mutual inductance?

A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 335 kV. The secondary of this transformer is being replaced so that its output can be 750 kV for more efficient cross-country transmission on upgraded transmission lines.

(a) What is the ratio of turns in the new secondary compared with the old secondary?

(b) What is the ratio of new current output to old output (at 335 kV) for the same power? (c) If the upgraded transmission lines have the same resistance, what is the ratio of new line power loss to old?

(a) Find the current through a \(0.500{\rm{ }}H\) inductor connected to a \(60.0{\rm{ }}Hz,{\rm{ }}480{\rm{ }}V\) AC source. (b) What would the current be at \(100{\rm{ }}kHz\)?

(a) Use the exact exponential treatment to find how much time is required to bring the current through an \({\rm{80}}{\rm{.0 mH}}\) inductor in series with a \({\rm{15}}{\rm{.0 \Omega }}\) resistor to \({\rm{99}}{\rm{.0\% }}\)of its final value, starting from zero. (b) Compare your answer to the approximate treatment using integral numbers of \({\rm{\tau }}\). (c) Discuss how significant the difference is.
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