Answer:
![B_T=2.0*10^-5[-\hat{i}+\hat{j}]T](https://tex.z-dn.net/?f=B_T%3D2.0%2A10%5E-5%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5DT)
Explanation:
To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current = 6.0 A
r: distance to the wire in which magnetic field is measured
In this case, you have four wires at corners of a square of length 9.0cm = 0.09m
You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.
If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:
![B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]](https://tex.z-dn.net/?f=B_T%3DB_1%2BB_2%2BB_3%5C%5C%5C%5CB_%7BT%7D%3D%5Cfrac%7B%5Cmu_o%20I_1%7D%7B2%5Cpi%20r_1%7D%5Chat%7Bj%7D-%5Cfrac%7B%5Cmu_o%20I_2%7D%7B2%5Cpi%20r_2%7D%5Chat%7Bi%7D%2B%5Cfrac%7B%5Cmu_o%20I_3%7D%7B2%5Cpi%20r_3%7D%5B-cos45%5Chat%7Bi%7D%2Bsin45%5Chat%7Bj%7D%5D)
I1 = I2 = I3 = 6.0A
r1 = 0.09m
r2 = 0.09m
Then you have:
![B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T](https://tex.z-dn.net/?f=B_T%3D%5Cfrac%7B%5Cmu_o%20I%7D%7B2%5Cpi%7D%5B%28-%5Cfrac%7B1%7D%7Br_2%7D-%5Cfrac%7Bcos45%7D%7Br_3%7D%29%5Chat%7Bi%7D%2B%28%5Cfrac%7B1%7D%7Br_1%7D%2B%5Cfrac%7Bsin45%7D%7Br_3%7D%29%5Chat%7Bj%7D%7D%5D%5C%5C%5C%5CB_T%3D%5Cfrac%7B%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%286.0A%29%7D%7B2%5Cpi%7D%5B%28-%5Cfrac%7B1%7D%7B0.09m%7D-%5Cfrac%7Bcos45%7D%7B0.127m%7D%29%5Chat%7Bi%7D%2B%28%5Cfrac%7B1%7D%7B0.09m%7D%2B%5Cfrac%7Bsin45%7D%7B0.127m%7D%29%5D%5C%5C%5C%5CB_T%3D%5Cfrac%7B%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%286.0A%29%7D%7B2%5Cpi%7D%5B-16.67%5Chat%7Bi%7D%2B16.67%5Chat%7Bj%7D%5D%5C%5C%5C%5CB_T%3D2.0%2A10%5E-5%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5DT)
Answer:
- The velocity component in the flow direction is much larger than that in the normal direction ( A )
- The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )
Explanation:
For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction
assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that
Vy << Vx
Answer: 3.12 * 10^12 F ( 3.12 pF)
Explanation: To calculate this capacitor of two hollow, coaxial, iron cylinders, we have to determine the potental differente between them and afeter that to use C=Q/ΔV
The electric field in th eregion rinner<r<router
By using the Gaussian law
∫E*ds=Q inside/εo
E*2*π*rinner^2*L= Q /εo
E=Q/(2*π*εo*r^2)
[Vab]=\int\limits^a_b {E} \, dr
where a and b are the inner and outer radii.
Then we have:
ΔV= 2*k*(Q/L)* ln (b/a)
replacing the values and using that C=Q/ΔV
we have:
C= L/(2*k*ln(b/a)=0.17/(2*9*10^9*3.023)=3.12 pF
Answer:
x₁ = 0.62 m
Explanation:
In this exercise the force is electric, given by Coulomb's law
F =
This force is a vector, since the three charges are in a line we can reduce the vector sum to a scalar sum.
For the sense of force let us use that charges of the same sign repel and charges of the opposite sign attract.
∑ F = F₁₂ - F₂₃
They ask us to find the point where the summaries of the force is zero.
F₁₂ - F₂₃ = 0
F₁₂ = F₂₃
let's fix a reference system located in the first charge (more to the left), the distance between the two charges is d = 1.5 m and x is the distance to the location of the second sphere
k q₁q₂ / x² = k q₂q₃ / (d-x) ²
q₁ (d-x) ² = q₃ x²
let's solve
d² - 2 x d + x² = 
x²
x² (1 - ) - 2x d + d² = 0
we substitute the values
x² (1- 4/2) - 2 1.5 x + 1.5² = 0
x² (-1) - 3.0 x + 2.25 = 0
x² + 3 x - 2.25 = 0
let's solve the quadratic equation
x = [-3 ± 
] / 2
x = [-3 ± 4.24] / 2
x₁ = 0.62 m
x₂ = 3.62 m
since it indicates that the charge q₂ e places between the spheres, the correct solution is
x₁ = 0.62 m
My estimate is 1/2 inch per week.
(0.5 inch/wk) x (1 wk / 7 da) x (1 da / 86,400 sec) x (1 meter / 39.37 inches)
= (0.5 x 1 x 1 x 1) / (7 x 86,400 x 39.37) meter/second
= 1 / 47,621,952 meter/second
= 0.0000000209987... m/s
= 2.1 x 10⁻⁸ meter/second
= 2.1 x 10⁻⁵ mm/sec
= 0.000021 millimeter per second (rounded)
