Problem#1
A 0.250-m-long bar moves on parallel rails that are connected through a 6.00 Ω resistor, as shown in Fig. 1, so the apparatus makes a complete circuit. You can ignore the resistance of the bar and rails. The circuit is in a uniform magnetic field B = 1.20 T that is directed into the plane of the figure. At an instant when the induced current in the circuit is counterclockwise and equal to 1.75 A, what is the velocity of the bar (magnitude and direction)? |
| Fig.1 |
Answer:
The motion of the bar causes an emf to be induced across its ends, which induces a current in the circuit.
F on the bar is to the left so v is to the right. Using ε = BvL and I = ε /R, we have
I = vBL/R,
So that,
v = IR/BL = (1.75 A)(6.0 Ω)/(1.20 T x 0.250 m) = 35.0 m/s
Problem#2
Measuring Blood Flow. Blood contains positive and negative ions and thus is a conductor. A blood vessel, therefore, can be viewed as an electrical wire. We can even picture the flowing blood as a series of parallel conducting slabs whose thickness is the diameter d of the vessel moving with speed v. (See Fig. 2.) (a) If the blood vessel is placed in a magnetic field B perpendicular to the vessel, as in the figure, show that the motional potential difference induced across it is ε = vBd. (b) If you expect that the blood will be flowing at 15 cm/s for a vessel 5.0 mm in diameter, what strength of magnetic field will you need to produce a potential difference of 1.0 mV? (c) Show that the volume rate of flow (R) of the blood is equal to R = πεd/4B. (Note: Although the method developed here is useful in measuring the rate of blood flow in a vessel, it is limited to use in surgery because measurement of the potential ε must be made directly across the vessel.)
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| Fig.2 |
Answer:
(a) Each slab of flowing blood has maximum width d and is moving perpendicular to the field with speed v. ε = vBL becomes ε = vBd.
(b) strength of magnetic field will you need to produce a potential difference of 1.0 mV is
B = ε/vd= 1.0 x 10-3 V/(0.15 m/s x 5.0 x 10-3 m) = 1.3 T
(c) The blood vessel has cross-sectional area
A = πd2/4
The volume of blood that flows past a cross section of the vessel in time t is
π(d2/4)vt
The volume flow rate is
volume/time = R = πd2v/4 and because v = ε/Bd, then
R = πdε/4B
Problem#3
A 1.41-m bar moves through a uniform, 1.20-T magnetic field with a speed of 2.50 m/s (Fig. 3). In each case, find the emf induced between the ends of this bar and identify which, if any, end (a or b) is at the higher potential. The bar moves in the direction of (a) the +x-axis, (b) the -y-axis, (c) the +z-axis, (d) How should this bar move so that the emf across its ends has the greatest possible value with b at a higher potential than a, and what is this maximum emf?
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| Fig.3 |
Answer:
Note that φ = 90° in all these cases because the bar moved perpendicular to the magnetic field. But the effective length of the bar, L sinθ, is different in each case.
(a) ε = vBL sinθ = (2.50 m/s)(1.20 T)(1.41 m) sin (37.0°) = 2.55 V, with a at the higher potential because positive charges are pushed toward that end.
(b) Same as (a) except θ = 53.0°, ε = (2.50 m/s)(1.20 T)(1.41 m) sin (53.0°) = 3.38 V, with a at the higher potential.
(c) Zero, since the velocity is parallel to the magnetic field.
(d) The bar must move perpendicular to its length, for which the emf is 4.23 V. For Vb > Va, it must move upward and to the left (toward the second quadrant) perpendicular to its length.
Problem#4
A rectangular circuit is moved at a constant velocity of 3.0 m/s into, through, and then out of a uniform 1.25-T magnetic field, as shown in Fig. 4. The magnetic-field region is considerably wider than 50.0 cm. Find the magnitude and direction (clockwise or counterclockwise) of the current induced in the circuit as it is (a) going into the magnetic field; (b) totally within the magnetic field, but still moving; and (c) moving out of the field. (d) Sketch a graph of the current in this circuit as a function of time, including the preceding three cases.
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| Fig.4 |
Answer:
(a) As the loop enters the field, the flux goes from zero to increasing out of the page. The induced magnetic field must counteract this, and points into the page and the current travels clockwise through the circuit. The magnitude of the induced emf is
ε = BLv
In terms of the current I, we also have ε = IR. Setting these equal gives
IR = BLv
Solving for I gives
I = BLv/R
I = (1.25 T)(0.75 m)(3.0 m/s)/(12.5 Ω)
I = 0.225 A
(b) Once the loop is entirely inside the magnetic field, there is no change in flux. If there is no flux, there is no induced magnetic field and therefore no current, so I = 0.
(c) As the loop leaves the field, the flux points out of the page but is decreasing. To balance this, the induced magnetic field points out of the page and the current travels counterclockwise through the circuit. Like in part (a), I = 0.225 A
(d) Let clockwise currents be positive. At t = 0 the loop is entering the field. It is totally in the field at time ta and beginning to move out of the field at time tb .The graph of the induced current as a function of time is sketched in Figure 5.
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| Fig.5 |
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