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Physics for Scientists and Engineers, 6e. Chapter 30 – Sources of the Magnetic Field.

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## Physics for Scientists and Engineers, 6e

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**Physics for Scientists and Engineers, 6e**Chapter 30 – Sources of the Magnetic Field**Consider the current in the length of wire shown in the**figure below. Rank the points A, B, and C, in terms of magnitude of the magnetic field due to the current in the length element shown, from greatest to least. • A, B, C • B, C, A • C, B, A • C, A, B • An equal field applies at all these points.**Point B is closest to the current element. Point C is**farther away and the field is further reduced by the sin θ factor in the cross product ds × . The field at A is zero because θ = 0.**For I1 = 2 A and I2 = 6 A in the figure below, which is**true: • F1 = 3F2 • F1 = F2/3 • F1 = F2**F1 = F2 as required by Newton’s third law. Another way to**arrive at this answer is to realize that Equation 30.11 gives the same result whether the multiplication of currents is (2 A)(6 A) or (6 A)(2 A).**A loose spiral spring is hung from the ceiling, and a large**current is sent through it. The coils move • closer together • farther apart • they do not move at all**The coils act like wires carrying parallel currents in the**same direction and hence attract one another.**Rank the magnitudes of B · ds for the closed paths in**the figure below, from least to greatest. • a, b, c, d • b, d, a, c • c, d, b, a • c, b, a, d • d, c, a, b**Equation 30.13 indicates that the value of the line integral**depends only on the net current through each closed path. Path b encloses 1 A, path d encloses 3 A, path a encloses 4 A, and path c encloses 6 A.**Rank the magnitudes of B · ds for the closed paths in**the figure below, from least to greatest. • a, b, c, d • b, c, d, a • b, d, a, c • d, c, a, b • The criteria is badly chosen in this case**Ranked from least to greatest, first comes b, then a = c =**d. Paths a, c, and d all give the same nonzero value μ0I because the size and shape of the paths do not matter. Path b does not enclose the current, and hence its line integral is zero.**Consider a solenoid that is very long compared to the**radius. Of the following choices, the most effective way to increase the magnetic field in the interior of the solenoid is: • double its length, keeping the number of turns per unit length constant • reduce its radius by half, keeping the number of turns per unit length constant • overwrapping the entire solenoid with an additional layer of current-carrying wire**The magnetic field in a very long solenoid is independent of**its length or radius. Overwrapping with an additional layer of wire increases the number of turns per unit length.**In an RC circuit, the capacitor begins to discharge. During**the discharge, in the region of space between the plates of the capacitor, there is • conduction current but no displacement current • displacement current but no conduction current • both conduction and displacement current • no current of any type**There can be no conduction current because there is no**conductor between the plates. There is a time-varying electric field because of the decreasing charge on the plates, and the time-varying electric flux represents a displacement current.**The capacitor in an RC circuit begins to discharge. During**the discharge, in the region of space between the plates of the capacitor, there is • an electric field but no magnetic field • a magnetic field but no electric field • both electric and magnetic fields • no fields of any type**There is a time-varying electric field because of the**decreasing charge on the plates. This time-varying electric field produces a magnetic field.**Which material would make a better permanent magnet?**• one whose hysteresis loop looks like Figure (a) • one whose hysteresis loop looks like Figure (b)**The loop that looks like Figure 30.32a is better because the**remanent magnetization at the point corresponding to point b in Figure 30.31 is greater.**If we wanted to cancel the Earth’s magnetic field by**running an enormous current loop around the equator, the current loop would be directed: • east to west • west to east**The lines of the Earth’s magnetic field enter the planet**in Hudson Bay and emerge from Antarctica; thus, the field lines resulting from the current would have to go in the opposite direction. Compare Figure 30.7a with Figure 30.36

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