Problem#1
A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth’s magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?Answer:
Length, L = 2.50 m
Current, I = 1.5 A
Magnetic field, B = 0.55 gauss = 0.55 x 10-4 T
(a) the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if is oriented so that the current in it is running from west to east is
FB = IL x B = ILB sin θ
FB = (1.5 A)(2.50 m)(0.55 x 10-4 T) sin 900
FB = 2.06 x 10-4 N, out of page
(b) the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if is oriented so that the current in it is running vertically upward is zero (there is no force) because angle between current and magnetic field is 00.
(c) the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if is oriented so that the current in it is running from north to south is zero (there is no force) because angle between current and magnetic field is 00.
(d) the largest possible force occurs when I is perpendicular to B (sin 900), as was the case in part (a). the force is a very small fraction of one newton, so it does not cause significant effects.
Problem#2
A straight, 2.00-m, 150-g wire carries a current in a region where the earth’s magnetic field is horizontal with a magnitude of 0.55 gauss. (a) What is the minimum value of the current in this wire so that its weight is completely supported by the magnetic force due to earth’s field, assuming that no other forces except gravity act on it? Does it seem likely that such a wire could support this size of current? (b) Show how the wire would have to be oriented relative to the earth’s magnetic field to be supported in this way.
Answer:
Length, L = 2.00 m
Mass, m = 150 g = 0.150 kg
Magnetic field, B = 0.55 gauss = 0.55 x 10-4 T
(a) The magnitude force is F = ILB sin θ. For the wire to be completely supported by the field requires that F = mg and that F and w are in opposite directions.
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| Fig.1 |
ILB = mg
I = mg/LB
I = (0.150 kg)(9.80 m/s2)/[(2.00 m)(0.55 x 10-4 T)] = 1.34 x 104 A.
This is a very large current and ohmic heating due to the resistance of the wire would be severe, such a current isn’t feasible.
(b) The magnetic force must upward. The directions of I, B, and F are shown Fig.1, where we have assumed that B is south to north. To produce an upward magnetic force, the current must be to the east. The wire must be horizontal and perpendicular to the earth’s magnetic field.
Problem#3
An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 cm between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force is exerted on the wire?
Answer:
Known:
Magnetic field, B = 0.550 T
Length, L = 2.50 cm = 0.0250 m is the length of wire in the magnetic field. Since the wire is perpendicular to B, θ = 900.
Current, i = 10.8 A
magnitude of force is exerted on the wire is
F = ILB sin θ
F = 10.8 A x 0.0250 m x 0.550 T
F = 0.1485 N
Problem#4
A long wire carrying 4.50 A of current makes two 900 bends, as shown in Fig. 2. The bent part of the wire passes through a uniform 0.240-T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.
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| Fig.2 |
Answer:
Known:
magnetic field, B = 0.240 T
current, I = 4.50 A
Figure 3a shows the direction of the force on each segment. Label the three segments in the field as P, Q and R. Let x be the length of segment P. Segment Q has length 30 cm and segment R has length 60 cm – x. For each segment θ = 900.
Known:
magnetic field, B = 0.240 T
current, I = 4.50 A
Figure 3a shows the direction of the force on each segment. Label the three segments in the field as P, Q and R. Let x be the length of segment P. Segment Q has length 30 cm and segment R has length 60 cm – x. For each segment θ = 900.
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| Fig.3 |
For segment P: FP = ILB = (4.50 A)(x)(0.240 T) = 1.08x
For segment Q: FQ = ILB = (4.50 A)(0.30 m)(0.240 T) = 0.324 N and is directed to the right.
For segment R, FR = ILB = (4.50 A)(0.60 – x)(0.240 T) = 1.08(0.60 – x)
Since FP and FR are in the same direction their vector sum has magnitude
FPR = FP + FR = 1.08x + 1.08(0.60 – x) = 0.648 N and is directed roward the bottom of the page in figure 3a.
The vector addition diagram for FPR and FQ is given in figure 3b.
F = [FPR2 + FQ2]1/2 = [(0.324 N)2 + (0.648 N)2]1/2 = 0.724 N
And
tan θ = FPR/FQ = 0.648/0.324, and θ = 63.40
so, the net force has magnitude 0.724 N and its direction is specified by θ = 63.40 in figure 3b.
Problem#5
A straight, vertical wire carries a current of 1.20 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude B = 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00-cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) 30.00 south of west?
Answer:
Known:
magnetic field, B = 0.588 T
current, I = 1.20 A
(a) The current and field direction are shows in Figure 4a. The right─hand rule that F is directed to the south, as shown.
F = ILB sin θ and θ = 900’
F = 1.20 A x 0.010 m x 0.588 T
F = 0.00706 N = 7.06 mN
For segment R, FR = ILB = (4.50 A)(0.60 – x)(0.240 T) = 1.08(0.60 – x)
Since FP and FR are in the same direction their vector sum has magnitude
FPR = FP + FR = 1.08x + 1.08(0.60 – x) = 0.648 N and is directed roward the bottom of the page in figure 3a.
The vector addition diagram for FPR and FQ is given in figure 3b.
F = [FPR2 + FQ2]1/2 = [(0.324 N)2 + (0.648 N)2]1/2 = 0.724 N
And
tan θ = FPR/FQ = 0.648/0.324, and θ = 63.40
so, the net force has magnitude 0.724 N and its direction is specified by θ = 63.40 in figure 3b.
Problem#5
A straight, vertical wire carries a current of 1.20 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude B = 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00-cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) 30.00 south of west?
Answer:
Known:
magnetic field, B = 0.588 T
current, I = 1.20 A
(a) The current and field direction are shows in Figure 4a. The right─hand rule that F is directed to the south, as shown.
F = ILB sin θ and θ = 900’
F = 1.20 A x 0.010 m x 0.588 T
F = 0.00706 N = 7.06 mN
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| Fig.4 |
(b) The right─hand rule gives that F is directed to the west, as shown in Figure 4b. θ = 900 and F = 7.06 mN, the same as in part 4(a)
(c) The current and field direction are shows in Figure 4c. The right─hand rule gives that F is 60.00 nort of west, θ = 900 and F = 7.06 mN, the same as in part 4(a)
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