EXAMPLE 21.1 Magnetic flux
For our first example, we use Equation 21.2 to solve for the strength of a magnetic field. A plane surface with area 3.0 cm^2 is placed in a uniform magnetic field that is oriented at an angle of 30° to the surface. (a) What is the angle phi? (b) If the magnetic flux through this area is 0.90 mWb, what is the magnitude of the magnetic field?
SOLUTION
SET UP AND SOLVE Part (a): Figure 21.6 shows our sketch. As defined earlier, phi is the angle between the direction of B and a line normal to the surface, so phi = 60° (not 30°).
Part (b): Because B and phi are the same at all points on the surface, we can use Equation 21.2: phi_B = BA cos phi. We solve for B, remembering to convert the area to square meters:
B = phi_B / (A cos phi) = (0.90 x 10^-3 Wb) / ((3.0 x 10^-4 m^2)(cos 60°)) = 6.0 T.
REFLECT The flux through the surface depends on its angle relative to B. Knowing this angle and the flux per unit area, we can find the magnitude of the magnetic field.
Practice Problem: For the same B and A, find the angle phi at which the flux would have half the value given above (i.e., the angle phi at which phi_B = 0.45 x 10^-3 Wb). Answer: 76°.
EXAMPLE 22.4 An R-L-C circuit
Now let's apply Equations 22.22 and 22.23 to a specific R-L-C circuit. The circuit layout is shown in Figure 22.13a. Suppose that its components have values R = 300 ohms, L = 60 mH, C = 0.50 microF, V = 50 V, and omega = 10,000 rad/s. Find the reactances XL, XC, and X, the impedance Z, the current amplitude I, the phase angle phi, and the voltage amplitude across each circuit element.
SOLUTION
SET UP If the circuit and phasor diagrams were not already provided in Figure 22.13, we would sketch them as the first step in this problem.
SOLVE From Equations 22.11 and 22.16, the reactances are
XL = omega*L = (10,000 rad/s)(60 x 10^-3 H) = 600 ohms,
XC = 1 / (omega*C) = 1 / ((10,000 rad/s)(0.50 x 10^-6 F)) = 200 ohms.
The reactance X of the circuit is
X = XL - XC = 600 ohms - 200 ohms = 400 ohms,
and the impedance Z is
Z = sqrt(R^2 + X^2) = sqrt((300 ohms)^2 + (400 ohms)^2) = 500 ohms.
With source voltage amplitude V = 50 V, the current amplitude I is
I = V / Z = 50 V / 500 ohms = 0.10 A.
The phase angle phi is
phi = arctan((XL - XC) / R) = arctan(400 ohms / 300 ohms) = 53°.
Because the phase angle phi is positive, the voltage leads the current by 53°. From Equation 22.6, the voltage amplitude VR across the resistor is
VR = IR = (0.10 A)(300 ohms) = 30 V.
From Equation 22.12, the voltage amplitude VL across the inductor is
VL = IXL = (0.10 A)(600 ohms) = 60 V.
From Equation 22.17, the voltage amplitude VC across the capacitor is
VC = IXC = (0.10 A)(200 ohms) = 20 V.
REFLECT Note that because of the phase differences between voltages across the separate elements, the source voltage amplitude V = 50 V is not equal to the sum of the voltage amplitudes across the separate circuit elements; that is, 50 V != 30 V + 60 V + 20 V. These voltages must be combined by vector addition of the corresponding phasors, not by simple numerical addition.
In this problem, the phase angle phi is positive, so the voltage leads the current by an angle (between 0 and 90°) equal to phi. If phi had turned out to be negative, then the voltage would lag the current by that angle.
Practice Problem: In a series circuit, suppose R = 100 ohms, L = 200 mH, C = 0.60 microF, V = 60 V, and omega = 4000 rad/s. Find the reactances XL and XC, the impedance Z, the current amplitude I, the phase angle phi, and the amplitude across each circuit element. Answers: XL = 800 ohms, XC = 420 ohms, Z = 400 ohms, I = 0.15 A, phi = 75°, VR = 15 V, VL = 120 V, VC = 63 V.