Engineering Electromagnetics, W. H. Hayt and J. A. Buck, McGraw Hill, 2001 (6th edition) 10장 족보
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Engineering Electromagnetics, W. H. Hayt and J. A. Buck, McGraw Hill, 2001 (6th edition) 연습문제 풀이 10장 풀이.
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10.1. In Fig. 10.4, let B = 0.2 cos 120πt T, and assume that the conductor joining the two ends
of the resistor is perfect. It may be assumed that the magnetic field produced by I(t) is
negligible. Find:
a) Vab(t): Since B is constant over the loop area, the flux is Φ = π(0.15)2B = 1.41 ×
10−2 cos 120πt Wb. Now, emf = Vba(t) = −dΦ/dt = (120π)(1.41 × 10−2) sin 120πt.
Then Vab(t) = −Vba(t) = −5.33 sin 120πt V.
b) I(t) = Vba(t)/R = 5.33 sin(120πt)/250 = 21.3 sin(120πt) mA
10.2. In Fig. 10.1, replace the voltmeter with a resistance, R.
a) Find the current I that flows as a result of the motion of the sliding bar: The current is
found through
I =
1
R
E · dL = − 1
R
dΦm
dt
Taking the normal to the path integral as az, the path direction will be counter-clockwise
when viewed from above (in the −az direction). The minus sign in the equation indicates
that the current will therefore flow clockwise, since the magnetic flux is increasing with
time. The flux of B is Φm = Bdvt, and so
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