Problem#1
A Changing Magnetic Field. You are testing a new data-acquisition system. This system allows you to record a graph of the current in a circuit as a function of time. As part of the test, you are using a circuit made up of a 4.00-cm-radius, 500-turn coil of copper wire connected in series to a resistor. Copper has resistivity 1.72 x 10-8 Ω.m and the wire used for the coil has diameter 0.0300 mm. You place the coil on a table that is tilted 30.00 from the horizontal and that lies between the poles of an electromagnet. The electromagnet generates a vertically upward magnetic field that is zero for t < 0 equal to (0.120 T) x (1 – cos πt) for 0 ≤ t ≤ 1.00 s and equal to 0.240 T for t > 1.00 s. (a) Draw the graph that should be produced by your data-acquisition system. (This is a full-featured system, so the graph will include labels and numerical values on its axes.) (b) If you were looking vertically downward at the coil, would the current be flowing clockwise or counterclockwise?Answer:
(a) The angle φ between the normal to the coil and the direction of B is 30.00.
|ε|= |dΦB/dt|
with ΦB = BA cos φ
ε = (Nπr2)(cos φ)(dB/dt)
for t < 0 and t > 1.00 s, dB/dt = 0, |ε|= 0 and I = 0.
For 0 ≤ t ≤ 1.00 s, dB/dt = (0.120 T/s)π sin πt
|ε| = (Nπr2)(cos φ)(0.120 T/s)π sin πt = (0.8206 V) sin πt
R for wire:
Rw = ρL/A
With, L = N(2πr) = 500 x 2π x (0.0400 m) = 125.7 m
Rw = ρL/πr2
Rw = (1.72 x 10-8 Ω.m)(125.7 m)/π(0.0150 x 10-3 m)2 = 3058 Ω
and the total resistance of the circuit is R = 3058 Ω + 600 Ω = 3658 Ω
I = ε/R = [(0.8206 V) sin πt]/3658 Ω
I = (0.224 mA) sin πt
The graph of I versus t is sketched in Figure 1a.
(b) The coil and the magnetic field are shown in Figure 1b.
B increasing so ΦB is ⊙ and increasing. Φind is ⊗ so is I clockwise
The long length of small diameter wire used to make the coil has a rather large resistance, larger than the resistance of the 600-Ω resistor connected to it in the circuit. The flux has a cosine time dependence so the rate of change of flux and the current have a sine time dependence. There is no induced current for t < 0 or t > 1.00 s.
Problem#2
A very long, straight solenoid with a crosssectional area of is wound with 90.0 turns of wire per centimeter. Starting at , the current in the solenoid is increasing according to i(t) = (0.160 A/s2)t2 . A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.20 A?
Answer:
The changing current in the solenoid will cause a changing magnetic field (and hence changing flux) through the secondary winding, which will induce an emf in the secondary coil.
The magnetic field of the solenoid is
B = µ0ni = (4π x 10-7 T.m/A)(90.0 x 102/m)(0.160 A/s2)t2
B = (1.810 x 10-3 T/s2)t2.
The total flux through secondary winding is
ΦB = NBA = 5 x (1.810 x 10-3 T/s2)t2 x (2.00 x 10-4 m2) = (1.810 x 10-6 Wb/s2)t2
Then,
|ε|= N|dΦB/dt|
|ε|= (3.619 x 10-6 V/s)t
the current in the solenoid is increasing according to
i(t) = (0.160 A/s2)t2, at t = 3.20 A, then
3.20 A = (0.160 A/s2)t2
t = 4.472 s
This gives
|ε|= (3.619 x 10-6 V/s)(4.472 s) = 1.62 x 10-5 V
Problem#3
A flat coil is oriented with the plane of its area at right angles to a spatially uniform magnetic field. The magnitude of this field varies with time according to the graph in Fig. 2. Sketch a qualitative (but accurate!) graph of the emf induced in the coil as a function of time. Be sure to identify the times t1, t2 and t3 on your graph.
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| Fig.2 |
A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil.
Faraday’s law says that the induced emf is
ε = -dΦB/dt
and the magnetic flux through a coil is defined as
ΦB = BA cos φ
In this case, , ΦB = BA where A is constant. So the emf is proportional to the negative slope of the magnetic field. The result is shown in Figure 3.
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| Fig.3 |
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