Q1. The inductor in Fig. E30.9 has inductance 0.260 H and carries a current in the direction shown that is decreasing at a uniform rate, . (a) Find the self-induced emf. (b) Which end of the inductor, a or b, is at a higher potential?

Background

Topic: Inductors and Electromagnetic Induction

This question tests your understanding of how an inductor responds to a changing current, specifically how to calculate the self-induced emf and determine the relative potential at the ends of the inductor.

Key Terms and Formulas

• Inductance (): A measure of an inductor's ability to oppose changes in current.

• Self-induced emf (): The emf generated within the inductor due to a changing current.

• Key formula:

• = inductance (in henries, H)

• = rate of change of current (in A/s)

• The negative sign indicates the direction of the induced emf opposes the change in current (Lenz's Law).

Step-by-Step Guidance

1. Identify the known values: , .

2. Write the formula for the self-induced emf: .

3. Substitute the given values into the formula: .

4. Consider the sign: Since is negative (current is decreasing), the induced emf will be positive, which means it opposes the decrease.

5. To determine which end (a or b) is at a higher potential, recall that the induced emf acts to maintain the current. The end where current enters the inductor will be at a higher potential when the current is decreasing.

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