Answer:
1.07 nT
Explanation:
We know that E/B = c where E = electric field amplitude = 320 mV/m = 0.32 V/m, B = magnetic field amplitude and c = speed of light = 3 × 10⁸ m/s.
So, B = E/c
Substituting E and c into B, we have
B = E/c
= 0.32 V/m ÷ 3 × 10⁸ m/s
= 0.1067 × 10⁻⁸ T
= 1.067 × 10⁻⁹ T
= 1.067 nT
≅ 1.07 nT
Make sure the triple beam balance is at 0 before you begin.
Answer:
Resistivity ![\rho =\frac{RA}{l}](https://tex.z-dn.net/?f=%5Crho%20%3D%5Cfrac%7BRA%7D%7Bl%7D)
It depends upon cross sectional area and length of material
Explanation:
The resistance of any material is given by
, here
is the resistivity of material , l is length of material and A is cross sectional area
So resistivity ![\rho =\frac{RA}{l}](https://tex.z-dn.net/?f=%5Crho%20%3D%5Cfrac%7BRA%7D%7Bl%7D)
So resistuivity of any material depends upon area of cross section and length of material
If cross sectional area will be more then resistivity will be more. And is length of the material will be more then resistivity will be less
The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as
![L = \frac{\mu N^2 A}{l}](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7B%5Cmu%20N%5E2%20A%7D%7Bl%7D)
Here,
= Permeability at free space
N = Number of loops
A = Cross-sectional Area
l = Length
Replacing with our values we have,
![L = \frac{(4\pi *10^{-7})(153)^2(\pi (\frac{5.11*10^{-3}}{2})^2)}{2.47*10^{-2}}](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7B%284%5Cpi%20%2A10%5E%7B-7%7D%29%28153%29%5E2%28%5Cpi%20%28%5Cfrac%7B5.11%2A10%5E%7B-3%7D%7D%7B2%7D%29%5E2%29%7D%7B2.47%2A10%5E%7B-2%7D%7D)
![L = 0.00002442H](https://tex.z-dn.net/?f=L%20%3D%200.00002442H)
![L = 24.42\mu H](https://tex.z-dn.net/?f=L%20%3D%2024.42%5Cmu%20H)
Therefore the Inductance is ![24.42\mu H](https://tex.z-dn.net/?f=24.42%5Cmu%20H)