Answer:
Good question to ask in physics, sir maam
Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;
![B = \frac{\mu I}{2 \pi r}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu%20I%7D%7B2%20%5Cpi%20r%7D)
Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire
![Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT](https://tex.z-dn.net/?f=Let%20%5C%20%5Cfrac%7B%5Cmu%20I%7D%7B2%5Cpi%7D%20%20%5C%20be%20%5C%20constant%3B%20%3D%20K%5C%5C%5C%5CB%20%3D%20%5Cfrac%7BK%7D%7Br%7D%20%5C%5C%5C%5CK%20%3D%20Br%5C%5C%5C%5CB_1r_1%20%3D%20B_2r_2%5C%5C%5C%5CB_2%20%3D%5Cfrac%7BB_1r_1%7D%7Br_2%7D%20%5C%5C%5C%5CB_2%20%3D%20%5Cfrac%7B2.5%2A10%5E%7B-3%7D%20%2A0.126%7D%7B0.198%7D%20%5C%5C%5C%5CB_2%20%3D%201.591%20%2A10%5E%7B-3%7D%5C%20T%5C%5C%5C%5CB_2%20%3D%201.591%20%5C%20mT)
Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
The time for the echo to return is directly proportional to the distance. vw = fλ. In a given medium under fixed conditions, vw is constant, so that there is a relationship between f and λ; the higher the frequency, the smaller the wavelength.
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S
= 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S
> σ