Respuesta :
Answer:
3.33 x 10⁻⁵ T
Explanation:
given,
current carried by two parallel wire = 25 A
distance between two wire = 13 cm
1) Let p be the distance between two wire
p = 13 cm = 0.13 m
a = 10 cm = 0.1 m
b = 6 cm = 0.06 m
using cosine law
[tex]p^2 = a^2 + b^2- 2ab cos\theta[/tex]
[tex]cos\theta =\dfrac{a^2+b^2}{2ab}[/tex]
[tex]\theta =cos^{-1}(\dfrac{0.1^2+0.06^2}{2\times 0.1 \times 0.06})[/tex]
θ = 105.96°
θ = 106°(approximately)
now, magnetic field on one wire
[tex]B_1 = \dfrac{\mu_oI}{2\pi r}[/tex]
[tex]B_1 = \dfrac{4\pi \times 10^{-7}\times 25}{2\pi (0.1)}[/tex]
B₁ = 5 x 10⁻⁵ T
now, magnetic field on another wire
[tex]B_2= \dfrac{\mu_oI}{2\pi r}[/tex]
[tex]B_2 = \dfrac{4\pi \times 10^{-7}\times 25}{2\pi (0.06)}[/tex]
B₂ = 8.33 x 10⁻⁵ T
resultant field,
= [tex]\sqrt{B_1^2+B_2^2 - 2 B_1B_2 cos \theta}[/tex]
= [tex]\sqrt{(5\times 10^{-5})^2+(8.33\times 10^{-5})^2 - 2\times 5 \times 10^{-5}\times 8.33 \times 10^{-5} cos 106^0}[/tex]
= 3.33 x 10⁻⁵ T
1) The magnitude of the magnetic field vector at the given point is; -3.195 * 10⁻⁵ T
2) The direction of the magnetic field vector at the given point is; 129.71° from the x-axis
What is the magnitude and direction of the magnetic Field?
We will start with finding the direction of the field for each wire from the tangent to the circle around the wire.
For their magnitudes, we have;
B₁ = (µ₀/4π) * 2I₁/L₁
B₁ = (4π * 10⁻⁷/4π) * 2 * 25/0.1
B₁ = 5 * 10⁻⁵ T
B₂ = (µ₀/4π) * 2I₂/L₂
B₂ = (4π * 10⁻⁷/4π) * 2 * 25/0.06
B₂ = 8.33 * 10⁻⁵ T
If we draw a vector diagram, we will have;
B = B₁(-sin θ₁ i + cos θ₁ j) + B₂(-sin θ₂ i + cos θ₂ j)
B = (5 * 10⁻⁵ )(-sin 26.34 i + cos 26.34 j) + (8.33 * 10⁻⁵)(-sin 47.7 i + cos 47.7 j)
B = -(8.38 * 10⁻⁵) i + (10.09 * 10⁻⁵) j
Direction of the field is;
θ = tan⁻¹((10.09 * 10⁻⁵)/-(8.38 * 10⁻⁵)
θ = 129.71° from the x-axis
Thus, magnitude is;
B_mag = B₁ * cos 129.71°
B_mag = -3.195 * 10⁻⁵ T
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