   Chapter 31, Problem 8P ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
ISBN: 9781337553278

#### Solutions

Chapter
Section ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
ISBN: 9781337553278
Textbook Problem

# The current in a 4.00 mH-inductor varies in time as shown in Figure P31.8. Construct a graph of the self-induced emf across the inductor over the time interval t = 0 to t = 12.0 ms. To determine

To draw: The graph of the self induced emf across the conductor for the given time interval.

Explanation

Explanation

Given Info: The inductance of the material is 4.00mH the time interval is from t=0s to t=40s .

The formula to calculate induced emf is,

ε=Ldidt

Here,

ε is the induced emf.

L is the inductance.

didt is the rate of change of current.

Tabulate the variation of current with respect to time according to given graph.

 t dt ms di(i2−i1) mA L mH ε=−Ldidt mV 0 to 2 ms 2 ms −1 − 0 = −1 4 2 2 to 4 ms 2

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