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Chapter 4: Q10DQ (page 947)

What are the relative advantages and disadvantages of Ampere’s law and the law of Biot-Savart for practical calculations of magnetic fields?

Short Answer

Expert verified

The advantage of Ampere’s law helps in finding electric field distribution and disadvantages is complicated for calculation.

The advantage of Biot-Savart law helps for calculate the field for something regardless of shape.

Step by step solution

01

Definition of Magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

02

Explain advantage and disadvantage

The advantage of Ampere’s law helps in finding electric field distribution and disadvantages is complicated for calculation.

The advantage of Biot-Savart law helps for calculates the field for something regardless of shape.

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Most popular questions from this chapter

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

A 10.0cm long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

When a resistor with resistance Ris connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that Rremains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

An idealized ammeter is connected to a battery as shown in Fig.

E25.28. Find (a) the reading of the ammeter, (b) the current through the4.00Ω

resistor, (c) the terminal voltage of the battery.

Fig. E25.28.

A 12.4-µF capacitor is connected through a 0.895-MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for t between 0 and 20 s
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