Problem#1
In Fig. 1, part of a long insulated wire carrying current i = 5.78 mA is bent into a circular section of radius R = 1.89 cm. In unit-vector notation, what is the magnetic field at the center of curvature C if the circular section (a) lies in the plane of the page as shown and (b) is perpendicular to the plane of the page after being rotated 90° counterclockwise as indicated? |
| Fig.1 |
(a) the magnetic field at the center of curvature C if the circular section lies in the plane of the page as shown is magnetic field on circular wire and straight wire with current,
BC = μ0i/2R (out the page) + μ0i/2πR (out the page)
BC = [μ0i/2R][1 + 1/π]
= [(4π x 10─7 T.m/A)(5.78 x 10─3 A)/(2 x 1.89 x 10─2 m)][1 + 1/π]
BC = 2.53 x 10─7 T
(b) the magnetic field at the center of curvature C if the circular section is perpendicular to the plane of the page after being rotated 90° counterclockwise as indicated
 |
| Fig.1 |
BC1 is perpendicular to BC2, then so the resultant magnetic fields is
BC2 = BC12 + BC22
BC2 = (μ0i/2πR)2 + (μ0i/2R)2
BC = [μ0i/2R][1 + 1/π2]1/2
= [(4π x 10─7 T.m/A)(5.78 x 10─3 A)/(2 x 1.89 x 10─2 m)][1 + 1/π2]1/2
BC = 2.02 x 10─7 T
And the direction of the resultant magnetic field is obtained as
tan θ = BC1/BC2 = 1/π
θ = tan─1(1/π) ≈ 17.660
so, in unit-vector notation, the magnetic field is written as
BC = BCxi + BCyj
BC = (2.02 x 10─7 T) cos 17.660 i + (2.02 x 10─7 T) sin 17.660 j
BC = (1.92i + 0.61j) x 10─7 T
Problem#10
A wire with current i = 3.00A is shown in Fig. 3. Two semi-infinite straight sections, both tangent to the same circle, are connected by a circular arc that has a central angle θ and runs along the circumference of the circle. The arc and the two straight sections all lie in the same plane. If B = 0 at the circles center, what is θ?
Answer:
The field due to both straight wires is out of the page while the field due to the circular arc is into the page. We just have to make them cancel.
(2 semi-infinite wire fields) = (field due to a circular arc)
If you forgot the field due to a semi-infinite wire, you’re not alone. So did I. But I know I can arrive at the expression from the precious few things I do know cold, such as Ampere’s law. Let’s do the work of getting that expression now using Ampere’s law and some symmetry. Your book derives the field due to a semi-infinite straight wire using the Biot-Savart law. This is formal, but probably the worst way to do it by hand quickly. You should avoid using Biot-Savart at all costs, in favor of Ampere’s law, when possible. The field due to a semi-infinite wire is half that of the infinite wire. That case is easy to handle with Ampere’s law.
In the case of the infinite wire, Ampere’s law delivers
Bi = μ0i/2πR
Now B for the semi-infinite case has the same direction and half the magnitude
Bsi = μ0i/4πR
This is the B we will use for the wires. Remember
2Bsi = Barc
The B due to a circular arc with radius R subtending an angle θ is still
Bi = μ0iθ/4πR
Now we just enforce 2Bsi = Barc
2[μ0i/2πR] = μ0iθ/4πR
θ = 2 rad = 114.590
Problem#11
Figure 4a shows two wires, each carrying a current.Wire 1 consists of a circular arc of radius R and two radial lengths; it carries current i1 = 2.0 A in the direction indicated. Wire 2 is long and straight; it carries a current i2 that can be varied; and it is at distance R/2 from the center of the arc. The net magnetic field due to the two currents is measured at the center of curvature of the arc. Figure 4b is a plot of the component of in the direction perpendicular to the figure as a function of current i2. The horizontal scale is set by i2s = 1.00 A. What is the angle subtended by the arc?
Answer:
Using Biot-Savart law, the radial sections of the wire does not contribute to the magnetic field at the center of the curvature. Therefore, we need only consider two contributions to the magnetic field at the center, one from the curved section of the left wire, and the other from the straight wire.
Using right-hand rule, we conclude the curved section will result in a magnetic field out of the page. Therefore, in order to for this to be cancelled out by the contribution from the straight wire, i2 must be flowing down, as indicated.
To find first contribution (by the curved wire) of the magnetic field can be found using Biot Savart law as follows:
B1 = μ0i1φ/4πR1 (out the page)
while the second contribution (by the straight wire) of the magnetic field can be found using Ampere’s law as:
B2 = μ0i2/2πR2 (into the page)
Therefore:
Bnet = B1 (out the page) + B2 (into the page)
Bnet = μ0i1φ/4πR1 ─ μ0i2/2πR2
Bnet = (μ0/2π)(i1φ/2R1 ─ i2/R2)
from the graph when i2 = 0.5 A the net magnetic field B = 0, so that
0 = (μ0/2π)(i1φ/2R1 ─ 1/2R2)
i1φ/2R1 = 1/2R2
φ = R1/i1R2
φ = R/(2.0 A x ½ R) = 1.0 rad = 57.30
Problem#12
In Fig. 6a, wire 1 consists of a circular arc and two radial lengths; it carries current i1 = 0.50 A in the direction indicated. Wire 2, shown in cross section, is long, straight, and perpendicular to the plane of the figure. Its distance from the center of the arc is equal to the radius R of the arc, and it carries a current i2 that can be varied. The two currents set up a net magnetic field B at the center of the arc. Figure 6b gives the square of the field’s magnitude B2 plotted versus the square of the current i22 . The vertical scale is set by Bs2 = 10.0 x 10-10 T2. What angle is subtended by the arc?
Answer:
Using Biot-Savart law, the radial sections of the wire does not contribute to the magnetic field at the center of the curvature. Therefore, we need only consider two contributions to the magnetic field at the center, one from the curved section of the left wire, and the other from the straight wire.
Center of circular are (carries a current i1):
B1 = μ0i1φ/4πR = μ0φ/8πR
For infinite straight wire (carries a current i2):
B2 = μ0i2/2πR
The net B field:
Bnet2 = B12 + B22
B2 = (μ0φ/8πR)2 + (μ0i2/2πR)2
B2 = (μ0/2πR)2[(i2)2 + (φ/4)2]
From the equation of a straight line with gradient m and intercept c on the y-axis is
y = mx + c
with, y = B2, x = i22; m = (μ0/2πR)2 and c = (μ0φ/8πR)2
from the graph we get, B2 = 6.0 x 10-10 T2 when i22 = 1 A2, then
m = B2/i22 = 6.0 x 10-10 T2/1 A2
(μ0/2πR)2 = 6.0 x 10-10 T2/1 A2
[(4π x 10─7 T.m/A)/(2πR)]2 = 6.0 x 10-10 T2/1 A2
4 x 10─14/R2 = 6.0 x 10-10
2 x 10─7/R = √6 x 10-5
R = 8.16 x 10-3 m
from the graph when x = 0, y = 0, and c = 1.0 x 10-10 T2, so that
c = (μ0φ/8πR)2
φ = (8πR√c)/μ0 = 8π x 8.16 x 10-3 x 10-5)/4π x 10─7
φ = 1.632 rad ≈ 93.510
BC2 = BC12 + BC22
BC2 = (μ0i/2πR)2 + (μ0i/2R)2
BC = [μ0i/2R][1 + 1/π2]1/2
= [(4π x 10─7 T.m/A)(5.78 x 10─3 A)/(2 x 1.89 x 10─2 m)][1 + 1/π2]1/2
BC = 2.02 x 10─7 T
And the direction of the resultant magnetic field is obtained as
tan θ = BC1/BC2 = 1/π
θ = tan─1(1/π) ≈ 17.660
so, in unit-vector notation, the magnetic field is written as
BC = BCxi + BCyj
BC = (2.02 x 10─7 T) cos 17.660 i + (2.02 x 10─7 T) sin 17.660 j
BC = (1.92i + 0.61j) x 10─7 T
Problem#10
A wire with current i = 3.00A is shown in Fig. 3. Two semi-infinite straight sections, both tangent to the same circle, are connected by a circular arc that has a central angle θ and runs along the circumference of the circle. The arc and the two straight sections all lie in the same plane. If B = 0 at the circles center, what is θ?
 |
| Fig.3 |
The field due to both straight wires is out of the page while the field due to the circular arc is into the page. We just have to make them cancel.
(2 semi-infinite wire fields) = (field due to a circular arc)
If you forgot the field due to a semi-infinite wire, you’re not alone. So did I. But I know I can arrive at the expression from the precious few things I do know cold, such as Ampere’s law. Let’s do the work of getting that expression now using Ampere’s law and some symmetry. Your book derives the field due to a semi-infinite straight wire using the Biot-Savart law. This is formal, but probably the worst way to do it by hand quickly. You should avoid using Biot-Savart at all costs, in favor of Ampere’s law, when possible. The field due to a semi-infinite wire is half that of the infinite wire. That case is easy to handle with Ampere’s law.
In the case of the infinite wire, Ampere’s law delivers
Bi = μ0i/2πR
Now B for the semi-infinite case has the same direction and half the magnitude
Bsi = μ0i/4πR
This is the B we will use for the wires. Remember
2Bsi = Barc
The B due to a circular arc with radius R subtending an angle θ is still
Bi = μ0iθ/4πR
Now we just enforce 2Bsi = Barc
2[μ0i/2πR] = μ0iθ/4πR
θ = 2 rad = 114.590
Problem#11
Figure 4a shows two wires, each carrying a current.Wire 1 consists of a circular arc of radius R and two radial lengths; it carries current i1 = 2.0 A in the direction indicated. Wire 2 is long and straight; it carries a current i2 that can be varied; and it is at distance R/2 from the center of the arc. The net magnetic field due to the two currents is measured at the center of curvature of the arc. Figure 4b is a plot of the component of in the direction perpendicular to the figure as a function of current i2. The horizontal scale is set by i2s = 1.00 A. What is the angle subtended by the arc?
 |
| Fig.4 |
Using Biot-Savart law, the radial sections of the wire does not contribute to the magnetic field at the center of the curvature. Therefore, we need only consider two contributions to the magnetic field at the center, one from the curved section of the left wire, and the other from the straight wire.
Using right-hand rule, we conclude the curved section will result in a magnetic field out of the page. Therefore, in order to for this to be cancelled out by the contribution from the straight wire, i2 must be flowing down, as indicated.
 |
| Fig.5 |
B1 = μ0i1φ/4πR1 (out the page)
while the second contribution (by the straight wire) of the magnetic field can be found using Ampere’s law as:
B2 = μ0i2/2πR2 (into the page)
Therefore:
Bnet = B1 (out the page) + B2 (into the page)
Bnet = μ0i1φ/4πR1 ─ μ0i2/2πR2
Bnet = (μ0/2π)(i1φ/2R1 ─ i2/R2)
from the graph when i2 = 0.5 A the net magnetic field B = 0, so that
0 = (μ0/2π)(i1φ/2R1 ─ 1/2R2)
i1φ/2R1 = 1/2R2
φ = R1/i1R2
φ = R/(2.0 A x ½ R) = 1.0 rad = 57.30
Problem#12
In Fig. 6a, wire 1 consists of a circular arc and two radial lengths; it carries current i1 = 0.50 A in the direction indicated. Wire 2, shown in cross section, is long, straight, and perpendicular to the plane of the figure. Its distance from the center of the arc is equal to the radius R of the arc, and it carries a current i2 that can be varied. The two currents set up a net magnetic field B at the center of the arc. Figure 6b gives the square of the field’s magnitude B2 plotted versus the square of the current i22 . The vertical scale is set by Bs2 = 10.0 x 10-10 T2. What angle is subtended by the arc?
 |
| Fig.6 |
Using Biot-Savart law, the radial sections of the wire does not contribute to the magnetic field at the center of the curvature. Therefore, we need only consider two contributions to the magnetic field at the center, one from the curved section of the left wire, and the other from the straight wire.
Center of circular are (carries a current i1):
B1 = μ0i1φ/4πR = μ0φ/8πR
For infinite straight wire (carries a current i2):
B2 = μ0i2/2πR
The net B field:
Bnet2 = B12 + B22
B2 = (μ0φ/8πR)2 + (μ0i2/2πR)2
B2 = (μ0/2πR)2[(i2)2 + (φ/4)2]
From the equation of a straight line with gradient m and intercept c on the y-axis is
y = mx + c
with, y = B2, x = i22; m = (μ0/2πR)2 and c = (μ0φ/8πR)2
from the graph we get, B2 = 6.0 x 10-10 T2 when i22 = 1 A2, then
m = B2/i22 = 6.0 x 10-10 T2/1 A2
(μ0/2πR)2 = 6.0 x 10-10 T2/1 A2
[(4π x 10─7 T.m/A)/(2πR)]2 = 6.0 x 10-10 T2/1 A2
4 x 10─14/R2 = 6.0 x 10-10
2 x 10─7/R = √6 x 10-5
R = 8.16 x 10-3 m
from the graph when x = 0, y = 0, and c = 1.0 x 10-10 T2, so that
c = (μ0φ/8πR)2
φ = (8πR√c)/μ0 = 8π x 8.16 x 10-3 x 10-5)/4π x 10─7
φ = 1.632 rad ≈ 93.510
Post a Comment for "Magnetic Field Due to a Current in a Circular Arc of Wire Problems and Solutions 3"