Answer:
The resulting magnetic force on the wire is -1.2kN
Explanation:
The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is
F = I (L*B)
Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)
L x B =
= (0, 0, -80)
we can now solve
F = I (L x B) = I (-80)
F = -1200 kmN
F = -1200 kN * 10⁻³
F = -1.2kN
I think i used calulater and it gives me 47.5
To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.
Since the propagation occurs in an area of spherical figure we will have to
![I = \frac{P}{A}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BP%7D%7BA%7D)
![I = \frac{P}{4\pi r^2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BP%7D%7B4%5Cpi%20r%5E2%7D)
Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that
![I = \frac{100}{4\pi (2.5)^2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B100%7D%7B4%5Cpi%20%282.5%29%5E2%7D)
![I = 1.2738W/m^2](https://tex.z-dn.net/?f=I%20%3D%201.2738W%2Fm%5E2)
The relation between intensity I and ![E_{max}](https://tex.z-dn.net/?f=E_%7Bmax%7D)
![I = \frac{E_max^2}{2\mu_0 c}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE_max%5E2%7D%7B2%5Cmu_0%20c%7D)
Here,
= Permeability constant
c = Speed of light
Rearranging for the Maximum Energy and substituting we have then,
![E_{max}^2 = 2I\mu_0 c](https://tex.z-dn.net/?f=E_%7Bmax%7D%5E2%20%3D%202I%5Cmu_0%20c)
![E_{max}=\sqrt{2I\mu_0 c }](https://tex.z-dn.net/?f=E_%7Bmax%7D%3D%5Csqrt%7B2I%5Cmu_0%20c%20%7D)
![E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)](https://tex.z-dn.net/?f=E_%7Bmax%7D%20%3D%202%281.2738%29%284%5Cpi%2A10%5E%7B-7%7D%29%283%2A10%5E8%29)
![E_{max} = 30.982 V/m](https://tex.z-dn.net/?f=E_%7Bmax%7D%20%3D%2030.982%20V%2Fm)
Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,
![B_{max} = \frac{E_{max}}{c}](https://tex.z-dn.net/?f=B_%7Bmax%7D%20%3D%20%5Cfrac%7BE_%7Bmax%7D%7D%7Bc%7D)
![B_{max} = \frac{30.982 V /m}{3*10^8}](https://tex.z-dn.net/?f=B_%7Bmax%7D%20%3D%20%5Cfrac%7B30.982%20V%20%2Fm%7D%7B3%2A10%5E8%7D)
![B_{max} = 1.03275 *10{-7} T](https://tex.z-dn.net/?f=B_%7Bmax%7D%20%3D%201.03275%20%2A10%7B-7%7D%20T)
Therefore the maximum value of the magnetic field is ![B_{max} = 1.03275 *10{-7} T](https://tex.z-dn.net/?f=B_%7Bmax%7D%20%3D%201.03275%20%2A10%7B-7%7D%20T)
Answer:
It helps the answer look clean. It also makes the work easier to work with.
Explanation:
Instead of writing a lot of zeros, all you have to do is add exponents to the number to show how much the decimal moved.
Answer:
The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.