Problem#1
Two long, parallel conductors, separated by 10.0 cm, carry currents in the same direction. The first wire carries current I1 = 5.00 A and the second carries I2 = 8.00 A. (a) What is the magnitude of the magnetic field created by I1 at the location of I2? (b) What is the force per unit length exerted by I1 on I2? (c) What is the magnitude of the magnetic field created by I2 at the location of I1? (d) What is the force per length exerted by I2 on I1?Answer:
Two long, parallel conductors, separated by 10.0 cm, carry currents in the same direction, let both wires carry current in the x direction.
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| Fig.1 |
(a) the magnitude of the magnetic field created by I1 at the location of I2 given by
B = μ0I/2πr
B = (4π x 10─7 T.m/A)(5.00 A)/( 2π x 0.10 m) = 1.00 x 10─5 T
by using the right hand rule we get the direction B out of the page (towards the positive z─axis)
B = (1.00 x 10─5 T)k
(b) the force per unit length exerted by I1 on I2 given by
F = I2L x B
F = 8.00 A(Li) x (1.00 x 10─5 T)k
F/L = ─(8.00 x 10─5 N)j
F/L = 8.00 x 10─5 N/m, toward the first wire
(c) the magnitude of the magnetic field created by I2 at the location of I1 given by
B = μ0I/2πr
B = (4π x 10─7 T.m/A)(8.00 A)/(2π x 0.10 m) = 1.60 x 10─5 T
by using the right hand rule we get the direction B into of the page (towards the positive ─z─axis)
B = ─(1.60 x 10─5 T)k
(d) the force per unit length exerted by I1 on I2 given by
F = I1L x B
F = 5.00 A(Li) x (─1.60 x 10─5 T)k
F/L = (8.00 x 10─5 N)j
F/L = 8.00 x 10─5 N/m, towards the second wire
Problem#2
In Figure 2, the current in the long, straight wire is I1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries the current I2 = 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and l = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.
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| Fig.2 |
Answer:
By symmetry, we note that the magnetic forces on the right and left segments of the rectangle cancel. The net force on the vertical segments of the rectangle is
F = F1 + F2
With F1 = μ0i1i2L/[2π(a + c)] and F2 = μ0i1i2L/2πc, then
F = ─μ0i1i2L/[2π(a + c)] + μ0i1i2L/2πc
F = [μ0i1i2L/2π][1/c ─ 1/(a + c)]
= [(4π x 10-7 Tm/A)(5.00 A)(10.0 A)(0.45 m)/2π][1/0.10 m ─ 1/0.25 m]
F = 2.70 x 10─5 N
F = F1 + F2
With F1 = μ0i1i2L/[2π(a + c)] and F2 = μ0i1i2L/2πc, then
F = ─μ0i1i2L/[2π(a + c)] + μ0i1i2L/2πc
F = [μ0i1i2L/2π][1/c ─ 1/(a + c)]
= [(4π x 10-7 Tm/A)(5.00 A)(10.0 A)(0.45 m)/2π][1/0.10 m ─ 1/0.25 m]
F = 2.70 x 10─5 N
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| Fig.3 |
In unit vector notation, the net force on the loop due to i1 is
F = (2.70 x 10─5 N)j = 27.0 μNj
Problem#3
Two long, parallel wires are attracted to each other by a force per unit length of 320 μN/m when they are separated by a vertical distance of 0.500 m. The current in the upper wire is 20.0 A to the right. Determine the location of the line in the plane of the two wires along which the total magnetic field is zero.
Answer:
To attract, both currents must be to into of the page. The attraction is described by
F = I2LB sin 900 = I2L(μ0I1/2πr)
F/L = μ0I1I2/2πr
Then,
320 x 10─6 N/m = (4π x 10-7 Tm/A)(20.0 A)I2/(2π x 0.500 m)
I2 = 40.0 A
F = (2.70 x 10─5 N)j = 27.0 μNj
Problem#3
Two long, parallel wires are attracted to each other by a force per unit length of 320 μN/m when they are separated by a vertical distance of 0.500 m. The current in the upper wire is 20.0 A to the right. Determine the location of the line in the plane of the two wires along which the total magnetic field is zero.
Answer:
To attract, both currents must be to into of the page. The attraction is described by
F = I2LB sin 900 = I2L(μ0I1/2πr)
F/L = μ0I1I2/2πr
Then,
320 x 10─6 N/m = (4π x 10-7 Tm/A)(20.0 A)I2/(2π x 0.500 m)
I2 = 40.0 A
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| Fig.4 |
Suppose that x represents the zero distance of the field to the right of wire 1. Then
B = B1 + B2
0 = ─μ0I1/{2πx} + μ0I2/{2π(0.50 m ─ x)}
I1/x = I2/(0.50 ─ x)
20.0 A/x = 40.0 A/(0.50 ─ x)
0.50 ─ x = 2x
x = 0.167 m (right wire 1)
Problem#4
Three long wires (wire 1, wire 2, and wire 3) hang vertically. The distance between wire 1 and wire 2 is 20.0 cm. On the left, wire 1 carries an upward current of 1.50 A. To the right, wire 2 carries a downward current of 4.00 A. Wire 3 is located such that when it carries a certain current, each wire experiences no net force. Find (a) the position of wire 3, and (b) the magnitude and direction of the current in wire 3.
Answer:
Carrying oppositely directed currents, wires 1 and 2 repel each other. If wire 3 were between them, it would have tobrepel either 1 or 2, so the force on that wire could not be zero. If wire 3 were to the right of wire 2, it would feel a larger force exerted by 2 than that exerted by 1, so the total force on 3 could not be zero. Therefore wire 3 must be to the left of both other wires as shown. It must carry downward current so that it can attract wire 2.
B = B1 + B2
0 = ─μ0I1/{2πx} + μ0I2/{2π(0.50 m ─ x)}
I1/x = I2/(0.50 ─ x)
20.0 A/x = 40.0 A/(0.50 ─ x)
0.50 ─ x = 2x
x = 0.167 m (right wire 1)
Problem#4
Three long wires (wire 1, wire 2, and wire 3) hang vertically. The distance between wire 1 and wire 2 is 20.0 cm. On the left, wire 1 carries an upward current of 1.50 A. To the right, wire 2 carries a downward current of 4.00 A. Wire 3 is located such that when it carries a certain current, each wire experiences no net force. Find (a) the position of wire 3, and (b) the magnitude and direction of the current in wire 3.
Answer:
Carrying oppositely directed currents, wires 1 and 2 repel each other. If wire 3 were between them, it would have tobrepel either 1 or 2, so the force on that wire could not be zero. If wire 3 were to the right of wire 2, it would feel a larger force exerted by 2 than that exerted by 1, so the total force on 3 could not be zero. Therefore wire 3 must be to the left of both other wires as shown. It must carry downward current so that it can attract wire 2.
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| Fig.5 |
(a) For the equilibrium of wire 3 we have
F31/L = F32/L
μ0i1i3/2πr13 = μ0i2i3/2πr23
(1.50 A)/d = (4.0 A)/(20 cm + d)
1.5(20 + d) = 4d
d = 30/2.5 = 12.0 cm to the left of wire 1
(b) For the equilibrium of wire 1 we have
F13/L = F12/L
μ0i1i3/2πr13 = μ0i2i1/2πr12
(1.50 A)(i3)/(12.0 cm) = (4.0 A)(1.5 A)/(20 cm)
i3 = 12 x 4 A/20 = 2.40 A (down)
We know that wire 2 must be in equilibrium because the forces on it are equal in magnitude to the forces that it exerts on wires 1 and 3, which are equal because they both balance the equal-magnitude forces that 1 exerts on 3 and that 3 exerts on 1.
Problem#5
The unit of magnetic flux is named for Wilhelm Weber. The practical-size unit of magnetic field is named for Johann Karl Friedrich Gauss. Both were scientists at Göttingen, Germany. Along with their individual accomplishments, together they built a telegraph in 1833. It consisted of a battery and switch, at one end of a transmission line 3 km long, operating an electromagnet at the other end. (André Ampère suggested electrical signaling in 1821; Samuel Morse built a telegraph line between Baltimore and Washington in 1844.) Suppose that Weber and Gauss’s transmission line was as diagrammed in Figure 6. Two long, parallel wires, each having a mass per unit length of 40.0 g/m, are supported in a horizontal plane by strings 6.00 cm long. When both wires carry the same current I, the wires repel each other so that the angle θ between the supporting strings is 16.0°. (a) Are the currents in the same direction or in opposite directions? (b) Find the magnitude of the current.
Answer:
The separation between the wires is
r = 2x = 6.00 cm(sin8.00°) = 1.67 cm.
(a) Because the wires repel, the currents are in opposite directions.
(b) Because the magnetic force acts horizontally, (z─axis), with
FB = μ0I2L/2πr
F31/L = F32/L
μ0i1i3/2πr13 = μ0i2i3/2πr23
(1.50 A)/d = (4.0 A)/(20 cm + d)
1.5(20 + d) = 4d
d = 30/2.5 = 12.0 cm to the left of wire 1
(b) For the equilibrium of wire 1 we have
F13/L = F12/L
μ0i1i3/2πr13 = μ0i2i1/2πr12
(1.50 A)(i3)/(12.0 cm) = (4.0 A)(1.5 A)/(20 cm)
i3 = 12 x 4 A/20 = 2.40 A (down)
We know that wire 2 must be in equilibrium because the forces on it are equal in magnitude to the forces that it exerts on wires 1 and 3, which are equal because they both balance the equal-magnitude forces that 1 exerts on 3 and that 3 exerts on 1.
Problem#5
The unit of magnetic flux is named for Wilhelm Weber. The practical-size unit of magnetic field is named for Johann Karl Friedrich Gauss. Both were scientists at Göttingen, Germany. Along with their individual accomplishments, together they built a telegraph in 1833. It consisted of a battery and switch, at one end of a transmission line 3 km long, operating an electromagnet at the other end. (André Ampère suggested electrical signaling in 1821; Samuel Morse built a telegraph line between Baltimore and Washington in 1844.) Suppose that Weber and Gauss’s transmission line was as diagrammed in Figure 6. Two long, parallel wires, each having a mass per unit length of 40.0 g/m, are supported in a horizontal plane by strings 6.00 cm long. When both wires carry the same current I, the wires repel each other so that the angle θ between the supporting strings is 16.0°. (a) Are the currents in the same direction or in opposite directions? (b) Find the magnitude of the current.
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| Fig.6 |
The separation between the wires is
r = 2x = 6.00 cm(sin8.00°) = 1.67 cm.
(a) Because the wires repel, the currents are in opposite directions.
(b) Because the magnetic force acts horizontally, (z─axis), with
FB = μ0I2L/2πr
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| Fig.7 |
Then from Fig.7b, we get
tan (θ/2) = FB/Fg
tan (θ/2) = μ0I2L/(2πrmg)
I2 = 2πmgr tan (θ/2)/(μ0L)
I2 = 2π x (40.0 x 10─3 kg/m)(9.80 m/s2)(1.67 x 10─2 m)(tan 80)/(4π x 10-7 Tm/A)
I2 = 4596.84 A2
So, I = 67.8 A
Problem#6
A long, horizontal wire AB rests on the surface of a table and carries a current I. Horizontal wire CD is vertically above wire AB and is free to slide up and down on the two vertical metal guides C and D (Fig. 8). Wire CD is connected through the sliding contacts to another wire that also carries a current I, opposite in direction to the current in wire AB. The mass per unit length of the wire CD is λ. To what equilibrium height h will the wire CD rise, assuming that the magnetic force on it is due entirely to the current in the wire AB?
Answer:
The force on CD are shown in figure 9.
Currents in opposite directions so the force is repulsive and FB is upward, as shown.
With,
FB = μ0I2L/(2πh) and Fg = mg = λLg
Where h is the distance between the wires and λ is the mass per unit length of the wire CD.
the wire CD rises until the upward force FB due to the currents balances the downward force of gravity.
tan (θ/2) = FB/Fg
tan (θ/2) = μ0I2L/(2πrmg)
I2 = 2πmgr tan (θ/2)/(μ0L)
I2 = 2π x (40.0 x 10─3 kg/m)(9.80 m/s2)(1.67 x 10─2 m)(tan 80)/(4π x 10-7 Tm/A)
I2 = 4596.84 A2
So, I = 67.8 A
Problem#6
A long, horizontal wire AB rests on the surface of a table and carries a current I. Horizontal wire CD is vertically above wire AB and is free to slide up and down on the two vertical metal guides C and D (Fig. 8). Wire CD is connected through the sliding contacts to another wire that also carries a current I, opposite in direction to the current in wire AB. The mass per unit length of the wire CD is λ. To what equilibrium height h will the wire CD rise, assuming that the magnetic force on it is due entirely to the current in the wire AB?
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| Fig.8 |
The force on CD are shown in figure 9.
Currents in opposite directions so the force is repulsive and FB is upward, as shown.
With,
FB = μ0I2L/(2πh) and Fg = mg = λLg
Where h is the distance between the wires and λ is the mass per unit length of the wire CD.
the wire CD rises until the upward force FB due to the currents balances the downward force of gravity.
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| Fig.9 |
Then, then in the equilibrium we write
FB = Fg
μ0I2L/(2πh) = λLg
h = μ0I2/2πλg
FB = Fg
μ0I2L/(2πh) = λLg
h = μ0I2/2πλg
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