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Q60PE

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College Physics (Urone)
Found in: Page 862

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Short Answer

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time must the \(0.100{\rm{ }}A\) current through a \(2.00{\rm{ }}mH\) inductor be switched on or off to induce a \(500{\rm{ }}V\) emf?

In \(\Delta t = 4 \cdot {10^{ - 4}}{\rm{ }}ms\) the \(0.100{\rm{ }}A\) current through a \(2.00{\rm{ }}mH\) inductor must be switched on or off to induce a \(500{\rm{ }}V\) emf.

See the step by step solution

Step by Step Solution

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

  • Current in the inductor: \(0.100{\rm{ }}A\).
  • Self-inductance of the inductor: \(\begin{array}{c}2.00{\rm{ }}mH = \frac{{2.00}}{{1000}}\\ = 0.002{\rm{ }}H\end{array}\).
  • EMF of the inductor to be induced: \(500{\rm{ }}V\).

Step 3: Calculation for time –

The absolute value of the electromotive force produced when an inductor's current varies is given by –

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

It will take a certain amount of time for a certain current shift to occur in a certain inductance in order to generate a certain electromotive force –

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

After substituting the values –

\(\begin{align}{}\Delta t &= \frac{{0.002 \cdot 0.1}}{{500}}\\ &= 4 \cdot {10^{ - 7}}{\rm{ }}s\\ &= 4 \cdot {10^{ - 7}} \times 1000{\rm{ }}s\\ &= 4 \cdot {10^{ - 4}}{\rm{ }}ms\end{align}\)

Therefore, the value for time is obtained as \(\Delta t = 4 \cdot {10^{ - 4}}{\rm{ }}ms\).

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