Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The wire at (x,y) – (-15.5 cm, 0) carries current l1-3.4 A in the negative z-direction. The wire at (x,y) (15.5 cm, 0) carries current 12-0.5 A in the positive z-direction. The wire at (x.y) -(0, 26.8 cm) carries current 13 -5.2 A in the positive z-direction. 1What is Bx0,0), the x-component of the magnetic field produced by these three wires at the origin? Submit 2)What is 8yl0,0), the y-component of the magnetic field produced by these three wires at the origin? Submit 3)What is Fx(1), the x-component of the force exerted on a one meter length of the wire carrying current 1? 4) What is Fyl), the y-component of the force exerted on a one meter length of the wire carrying current 1? 5)What is Fx(2), the x-component of the force exerted on a one meter length of the wire carrying current l2? Submit 6)Another wire is now added, also aligned with the z-axis at (x,y) (0, -26.8 cm) as shown. This wire carries current 14 A. Which of the following statements is true If 14 is directed along the positive z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin If 14 is directed along the negative z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin If 14 is directed along the positive z-axis, then it is possible to make the x-component of the magnetic field equal to zero at the origin If 14 is directed along the negative z-axis, then it is possible to make the x- component of the magnetic field equal to zero at the origin
The Correct Answer and Explanation is:
Let’s solve the problem step by step, starting with the individual components for each question.
1) Bx(0,0), the x-component of the magnetic field produced by these three wires at the origin
The magnetic field due to a current-carrying wire at a distance rrr from the wire is given by Ampère’s law:B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}B=2πrμ0I
Where:
- BBB is the magnetic field,
- μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}μ0=4π×10−7T\cdotpm/A (permeability of free space),
- III is the current,
- rrr is the perpendicular distance from the wire to the point where the magnetic field is measured.
For the three wires, we will calculate the magnetic field at the origin (0,0) due to each wire.
- Wire 1 at (-15.5 cm, 0), I1=3.4 AI_1 = 3.4 \, \text{A}I1=3.4A, direction: negative z-axis: The magnetic field produced by a current-carrying wire has a direction given by the right-hand rule. Since this wire is to the left of the origin, the field at the origin will point in the negative yyy-direction.The distance r1=15.5 cm=0.155 mr_1 = 15.5 \, \text{cm} = 0.155 \, \text{m}r1=15.5cm=0.155m, and the magnetic field due to this wire at the origin is:B1=μ0I12πr1=(4π×10−7)×3.42π×0.155≈4.4×10−6 TB_1 = \frac{\mu_0 I_1}{2\pi r_1} = \frac{(4\pi \times 10^{-7}) \times 3.4}{2\pi \times 0.155} \approx 4.4 \times 10^{-6} \, \text{T}B1=2πr1μ0I1=2π×0.155(4π×10−7)×3.4≈4.4×10−6TSince this field points in the negative yyy-direction, the contribution to Bx(0,0)Bx(0, 0)Bx(0,0) is 0.
- Wire 2 at (15.5 cm, 0), I2=0.5 AI_2 = 0.5 \, \text{A}I2=0.5A, direction: positive z-axis: This wire is to the right of the origin. Using the right-hand rule, the magnetic field at the origin will point in the positive yyy-direction.The distance r2=0.155 mr_2 = 0.155 \, \text{m}r2=0.155m, and the magnetic field due to this wire at the origin is:B2=μ0I22πr2=(4π×10−7)×0.52π×0.155≈6.4×10−7 TB_2 = \frac{\mu_0 I_2}{2\pi r_2} = \frac{(4\pi \times 10^{-7}) \times 0.5}{2\pi \times 0.155} \approx 6.4 \times 10^{-7} \, \text{T}B2=2πr2μ0I2=2π×0.155(4π×10−7)×0.5≈6.4×10−7TSince the field is in the positive yyy-direction, the contribution to Bx(0,0)Bx(0,0)Bx(0,0) is 0.
- Wire 3 at (0, 26.8 cm), I3=5.2 AI_3 = 5.2 \, \text{A}I3=5.2A, direction: positive z-axis: This wire is directly above the origin. The magnetic field produced by this wire at the origin will point in the negative xxx-direction (by the right-hand rule).The distance r3=26.8 cm=0.268 mr_3 = 26.8 \, \text{cm} = 0.268 \, \text{m}r3=26.8cm=0.268m, and the magnetic field due to this wire at the origin is:B3=μ0I32πr3=(4π×10−7)×5.22π×0.268≈2.4×10−6 TB_3 = \frac{\mu_0 I_3}{2\pi r_3} = \frac{(4\pi \times 10^{-7}) \times 5.2}{2\pi \times 0.268} \approx 2.4 \times 10^{-6} \, \text{T}B3=2πr3μ0I3=2π×0.268(4π×10−7)×5.2≈2.4×10−6TThis field is in the negative xxx-direction, so it contributes to Bx(0,0)Bx(0,0)Bx(0,0) with a value of −2.4×10−6 T-2.4 \times 10^{-6} \, \text{T}−2.4×10−6T.
Thus, the x-component of the magnetic field at the origin is:Bx(0,0)=0+0−2.4×10−6=−2.4×10−6 TBx(0,0) = 0 + 0 – 2.4 \times 10^{-6} = -2.4 \times 10^{-6} \, \text{T}Bx(0,0)=0+0−2.4×10−6=−2.4×10−6T
2) By(0,0), the y-component of the magnetic field produced by these three wires at the origin
- Wire 1 contributes a magnetic field in the negative yyy-direction with magnitude 4.4×10−6 T4.4 \times 10^{-6} \, \text{T}4.4×10−6T.
- Wire 2 contributes a magnetic field in the positive yyy-direction with magnitude 6.4×10−7 T6.4 \times 10^{-7} \, \text{T}6.4×10−7T.
- Wire 3 contributes no magnetic field in the yyy-direction (it contributes only in the xxx-direction).
Thus, the y-component of the magnetic field at the origin is:By(0,0)=−4.4×10−6+6.4×10−7=−3.76×10−6 TBy(0,0) = -4.4 \times 10^{-6} + 6.4 \times 10^{-7} = -3.76 \times 10^{-6} \, \text{T}By(0,0)=−4.4×10−6+6.4×10−7=−3.76×10−6T
3) Fx(1), the x-component of the force exerted on a one meter length of wire carrying current I1
The force per unit length on a current-carrying wire in a magnetic field is given by:F/L=I⋅B⋅sin(θ)F/L = I \cdot B \cdot \sin(\theta)F/L=I⋅B⋅sin(θ)
Where:
- I=3.4 AI = 3.4 \, \text{A}I=3.4A,
- B=2.4×10−6 TB = 2.4 \times 10^{-6} \, \text{T}B=2.4×10−6T (from the magnetic field calculation),
- θ=90∘\theta = 90^\circθ=90∘ (since the magnetic field is perpendicular to the wire).
Thus, the force per unit length on wire 1 is:F1=3.4⋅2.4×10−6=8.16×10−6 N/mF_1 = 3.4 \cdot 2.4 \times 10^{-6} = 8.16 \times 10^{-6} \, \text{N/m}F1=3.4⋅2.4×10−6=8.16×10−6N/m
This force is directed in the positive yyy-direction (because wire 1’s current is in the negative zzz-direction and the magnetic field produced by wire 3 is in the negative xxx-direction).
So, the x-component of the force on wire 1 is 0, and the y-component of the force on wire 1 is 8.16×10−6 N/m8.16 \times 10^{-6} \, \text{N/m}8.16×10−6N/m.
4) Fy(1), the y-component of the force exerted on a one meter length of wire carrying current I1
The magnetic field at wire 1 is primarily in the yyy-direction, causing a force in the xxx-direction. Since the force on wire 1 is perpendicular to both the wire and the magnetic field, the yyy-component of the force is 0.
5) Fx(2), the x-component of the force exerted on a one meter length of the wire carrying current I2
For wire 2, the magnetic field at its location is in the yyy-direction. Since the current in wire 2 is in the positive zzz-direction, the force is directed in the positive xxx-direction.F2=0.5×2.4×10−6=1.2×10−6 N/mF_2 = 0.5 \times 2.4 \times 10^{-6} = 1.2 \times 10^{-6} \, \text{N/m}F2=0.5×2.4×10−6=1.2×10−6N/m
Thus, Fx(2)=1.2×10−6 N/mFx(2) = 1.2 \times 10^{-6} \, \text{N/m}Fx(2)=1.2×10−6N/m.
6) Effect of adding a new wire at (0, -26.8 cm)
Adding a new wire at (0,−26.8 cm)(0, -26.8 \, \text{cm})(0,−26.8cm) with a current I4=4 AI_4 = 4 \, \text{A}I4=4A will influence the magnetic field at the origin. The magnetic field produced by this wire will interact with the fields from the other wires.
- If I4I_4I4 is directed along the positive zzz-axis, it will create a magnetic field that increases the yyy-component, and it will be possible to make By(0,0)=0B_y(0,0) = 0By(0,0)=0.
- If I4I_4I4 is directed along the negative zzz-axis, it will create a magnetic field that decreases the yyy-component, potentially making it zero.
The correct answer is:
“If I4I_4I4 is directed along the positive z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin.”