Problem#1
A rectangular loop of wire with dimensions 1.50 cm by 8.00 cm and resistance R = 0.600 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude B = 3.50 T and is directed into the plane of Fig. 1. At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what force (magnitude and direction) does the magnetic field exert on the loop? |
| Fig.1 |
Answer:
The magnetic flux through the loop is decreasing, so an emf will be induced in the loop, which will induce a current in the loop.
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| Fig.2 |
The magnetic field will exert a force on the loop due to this current.
The motional ε is ε = vBL, I = ε/R, and FB = ILB, then
FB = (vBL/R)LB = vB2L2/R
FB = (3.00 m/s)(3.50 T)2(0.0150 m)2/0.600 Ω
FB = 0.0138 N
B is into the page and ΦB is decreasing, so the field of the induced current is into the page inside the loop and the induced current is clockwise. Using F = IL x B, we see that the force on the left-hand end of the loop to be to the left.
Problem#2
In Fig. 3 a conducting rod of length L = 30.0 cm moves in a magnetic field of magnitude 0.450 T directed into the plane of the figure. The rod moves with speed v = 5.00 m/s in the direction shown. (a) What is the potential difference between the ends of the rod? (b) Which point, a or b, is at higher potential? (c) When the charges in the rod are in equilibrium, what are the magnitude and direction of the electric field within the rod? (d) When the charges in the rod are in equilibrium, which point, a or b, has an excess of positive charge? (e) What is the potential difference across the rod if it moves (i) parallel to ab and (ii) directly out of the page?
 |
| Fig.3 |
A conductor moving in a magnetic field may have a potential difference induced across it, depending on how it is moving.
The induced emf is
ε = vBL sin θ,
where θ is the angle between the velocity and the magnetic field.
(a) the potential difference between the ends of the rod
ε = vBL sin θ = (5.00 m/s)(0.450 T)(0.30 m) sin 900
ε = 0.675 V
(b) The positive charges are moved to end b, so b is at the higher potential.
(c) the magnitude and direction of the electric field within the rod is
E = V/L = (0.675 V)/(0.300 m) = 2.25 V/m. The direction of E is from b to a.
(d) The positive charge are pushed to b, so b has an excess of positive charge.
(e) (i) If the rod has no appreciable thickness, L = 0, so the emf is zero. (ii) The emf is zero because no magnetic force acts on the charges in the rod since it moves parallel to the magnetic field.
Problem#3
A rectangle measuring 30.0 cm by 40.0 cm is located inside a region of a spatially uniform magnetic field of 1.25 T, with the field perpendicular to the plane of the coil (Fig. 4). The coil is pulled out at a steady rate of 2.00 cm/s traveling perpendicular to the field lines. The region of the field ends abruptly as shown. Find the emf induced in this coil when it is (a) all inside the field; (b) partly inside the field; (c) all outside the field.
 |
| Fig.4 |
Answer:
(a) The magnetic flux is constant, so the induced emf is zero (ε = 0).
(b) The area inside the field is changing. If we let x be the length (along the 30.0-cm side) in the field, then
A = (0.400 m)x
Then, ΦB = BA = (1.25 T)(0.400 m)x
ε = |dΦB/dt|= Bd[(0.400 m)x]/dt = B(0.400 m)(dx/dt) = B(0.400 m)v
So that
ε = (1.25 T)(0.400 m)(0.0200 m/s) = 0.0100 V
(c) The magnetic flux is constant, so the induced emf is zero (ε = 0).
Problem#4
Are Motional emfs a Practical Source of Electricity? How fast (in and mph) would a 5.00-cm copper bar have to move at right angles to a 0.650-T magnetic field to generate 1.50 V (the same as a AA battery) across its ends? Does this seem like a practical way to generate electricity?
Answer:
we use
ε = vBL, then
v = ε/BL = 1.50 V/(0.650 T x 0.0050 m) = 46.2 m/s or
v = 103 mph
This is a large speed and not practical. It is also difficult to produce a 5.00-cm wide region of 0.650 T magnetic field.
Problem#5
Motional emfs in Transportation. Airplanes and trains move through the earth’s magnetic field at rather high speeds, so it is reasonable to wonder whether this field can have a substantial effect on them. We shall use a typical value of 0.50 G for the earth’s field (a) The French TGV train and the Japanese “bullet train” reach speeds of up to 180 mph moving on tracks about 1.5 m apart. At top speed moving perpendicular to the earth’s magnetic field, what potential difference is induced across the tracks as the wheels roll? Does this seem large enough to produce noticeable effects? (b) The Boeing 747-400 aircraft has a wingspan of 64.4 m and a cruising speed of 565 mph. If there is no wind blowing (so that this is also their speed relative to the ground), what is the maximum potential difference that could be induced between the opposite tips of the wings? Does this seem large enough to cause problems with the plane?
Answer:
ε = vBL
1 mph = 0.4470 m/s
1 G = 10-4 T
(a) ε = (180 mph)(0.4770 m/s/1 mph)(0.50 x 10-4 T)(1.5 m) = 6.0 mV
This is much too small to be noticeable.
(b) ε = (565 mph)(0.4470 m/s/1 mph)(0.50 x 10-4 T)(64.4 m) = 0.813 mV
This is too small to be noticeable.
Even through the speeds and values of L are large, the earth’s field is small and motional emfs due to the earth’s field are not important in these situations.
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