Consider the rectangular (an amperian loop) path of length L and width w shown in Figure 5.2 (right). We can apply Ampère's law to this path by evaluating the integral of E-de over each side of the rectangle. The contribution along side 3 is zero because the magnetic field is zero outside the solenoid. The contributions from sides 2 and 4 are both zero, because is perpendicular to då along these paths, both inside and outside the solenoid Side 1 gives a contribution to the integral because along this path 5 is uniform and parallel to dž, that is, the only contribution comes from path 1. Blds Bids Blds B=0 fE dš = B đš + B-dš + B đš + B-d3 = BL = 4olenci path 1 path 2 path 3 path 4 If N is the mumber of tums in the length L, the total curent through the rectangle is NI. Therefore, Ampère's law applied to this path gives N B-ds = BL = H9NI =B = H0 = Honl 5.2 where n = N/L is the number of tums per unit length. Consider a section of the solenoid of length dx'. The total current winding around the solenoid in that section is dI = nidx'. This section is located at a distance x – x' away from the point P. The contribution to the magnetic field at P due to this subset of loops is HOR? 2[(x – x')² + R²]ª/2 HOR? dB = dl =- 5.3 2[(x - x')? + R2j3/2 nldx Integrating over the entire length of the solenoid, we obtain +L/2 |+L/2 dx' HonIR? (x – x') [(x – x')² + R°j³/2 R²[(x – x')² + R*Il_1/2 Honi x+L/2 x- L/2 5.4 2 Vx+L/2}² + R² V(x- L/2)? + R². 2 Equation 5.4 expresses the change of magnetic field depending on the distance from the center.
Consider the rectangular (an amperian loop) path of length L and width w shown in Figure 5.2 (right). We can apply Ampère's law to this path by evaluating the integral of E-de over each side of the rectangle. The contribution along side 3 is zero because the magnetic field is zero outside the solenoid. The contributions from sides 2 and 4 are both zero, because is perpendicular to då along these paths, both inside and outside the solenoid Side 1 gives a contribution to the integral because along this path 5 is uniform and parallel to dž, that is, the only contribution comes from path 1. Blds Bids Blds B=0 fE dš = B đš + B-dš + B đš + B-d3 = BL = 4olenci path 1 path 2 path 3 path 4 If N is the mumber of tums in the length L, the total curent through the rectangle is NI. Therefore, Ampère's law applied to this path gives N B-ds = BL = H9NI =B = H0 = Honl 5.2 where n = N/L is the number of tums per unit length. Consider a section of the solenoid of length dx'. The total current winding around the solenoid in that section is dI = nidx'. This section is located at a distance x – x' away from the point P. The contribution to the magnetic field at P due to this subset of loops is HOR? 2[(x – x')² + R²]ª/2 HOR? dB = dl =- 5.3 2[(x - x')? + R2j3/2 nldx Integrating over the entire length of the solenoid, we obtain +L/2 |+L/2 dx' HonIR? (x – x') [(x – x')² + R°j³/2 R²[(x – x')² + R*Il_1/2 Honi x+L/2 x- L/2 5.4 2 Vx+L/2}² + R² V(x- L/2)? + R². 2 Equation 5.4 expresses the change of magnetic field depending on the distance from the center.
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Transcribed Image Text:Questions:
1. Show that for L » R, Equation 5.4 is equivalent to Equation 5.2.
![Consider the rectangular (an amperian loop) path of length L and width w shown in Figure 5.2 (right).
We can apply Ampère's law to this path by evaluating the integral of E-de over each side of the rectangle.
The contribution along side 3 is zero because the magnetic field is zero outside the solenoid. The
contributions from sides 2 and 4 are both zero, because is perpendicular to då along these paths, both
inside and outside the solenoid Side 1 gives a contribution to the integral because along this path 5 is
uniform and parallel to dž, that is, the only contribution comes from path 1.
Blds
Bids
Blds
B=0
fE dš = B đš + B-dš + B đš + B-d3 = BL = 4olenci
path 1
path 2
path 3
path 4
If N is the mumber of tums in the length L, the total curent through the rectangle is NI. Therefore,
Ampère's law applied to this path gives
N
B-ds = BL = H9NI =B = H0 = Honl
5.2
where n = N/L is the number of tums per unit length.
Consider a section of the solenoid of length dx'. The total current winding around the solenoid in that
section is dI = nidx'. This section is located at a distance x – x' away from the point P. The contribution
to the magnetic field at P due to this subset of loops is
HOR?
2[(x – x')² + R²]ª/2
HOR?
dB =
dl =-
5.3
2[(x - x')? + R2j3/2 nldx
Integrating over the entire length of the solenoid, we obtain
+L/2
|+L/2
dx'
HonIR?
(x – x')
[(x – x')² + R°j³/2
R²[(x – x')² + R*Il_1/2
Honi
x+L/2
x- L/2
5.4
2 l
(x +L/2)2 +R2
V(x- L/2)² + R²
Equation 5.4 expresses the change of magnetic field depending on the distance from the center.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc386abc3-0420-4cc0-89bb-4a7032d67409%2F0ef61ac4-10f8-4171-94ff-2700427d1373%2Ft65k3pn_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the rectangular (an amperian loop) path of length L and width w shown in Figure 5.2 (right).
We can apply Ampère's law to this path by evaluating the integral of E-de over each side of the rectangle.
The contribution along side 3 is zero because the magnetic field is zero outside the solenoid. The
contributions from sides 2 and 4 are both zero, because is perpendicular to då along these paths, both
inside and outside the solenoid Side 1 gives a contribution to the integral because along this path 5 is
uniform and parallel to dž, that is, the only contribution comes from path 1.
Blds
Bids
Blds
B=0
fE dš = B đš + B-dš + B đš + B-d3 = BL = 4olenci
path 1
path 2
path 3
path 4
If N is the mumber of tums in the length L, the total curent through the rectangle is NI. Therefore,
Ampère's law applied to this path gives
N
B-ds = BL = H9NI =B = H0 = Honl
5.2
where n = N/L is the number of tums per unit length.
Consider a section of the solenoid of length dx'. The total current winding around the solenoid in that
section is dI = nidx'. This section is located at a distance x – x' away from the point P. The contribution
to the magnetic field at P due to this subset of loops is
HOR?
2[(x – x')² + R²]ª/2
HOR?
dB =
dl =-
5.3
2[(x - x')? + R2j3/2 nldx
Integrating over the entire length of the solenoid, we obtain
+L/2
|+L/2
dx'
HonIR?
(x – x')
[(x – x')² + R°j³/2
R²[(x – x')² + R*Il_1/2
Honi
x+L/2
x- L/2
5.4
2 l
(x +L/2)2 +R2
V(x- L/2)² + R²
Equation 5.4 expresses the change of magnetic field depending on the distance from the center.
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