Respuesta :
Answer:
[tex]r = 3.36 \times 10^{-7} m[/tex]
Explanation:
As per Ampere's law of magnetic field we know that
line integral of magnetic field along closed ampere's loop is equal to the product of current enclosed and magnetic permeability of medium
So it is given as
[tex]\int B. dl = \mu_0 i_{en}[/tex]
here we can say that enclosed current is given as
[tex]i_{en} = \frac{i}{\pi R^2} (\pi r^2)[/tex]
now from ampere'e loop law for any point inside the wire we will have
[tex]B.(2\pi r) = \mu_o (\frac{ir^2}{R^2}[/tex]
[tex]B = \frac{\mu_0 i r}{2\pi R^2}[/tex]
now we know that magnetic field inside the wire is 84% of the field at its surface
so we will have
[tex]0.84 \frac{\mu_o i}{2\pi R} = \frac{\mu_o i r}{2\pi R^2}[/tex]
so we have
[tex]r = 0.84 R[/tex]
now we know
[tex]\frac{\mu_o i}{2\pi R} = 1[/tex]
here i = 2 A
[tex]R = 2\times 10^{-7} m[/tex]
so now we have
[tex]r = 3.36 \times 10^{-7} m[/tex]
The point ( r ) within the wire where the field strength equals 84% of the field strength at the wire surface is : 0.84 R
Given data :
Radius of wire = R
current in the wire = 2A
magnetic field strength = 1 T
Determine the area within the wire where the field strength equals 84%
we will apply Ampere's law
i) Ampere's law applied inside the wire
B₁ (2πr ) = μ₀I ( r² / R² )
ii) Ampere's law applied at the surface
B₂ ( 2πr ) = μ₀ I
Resolving equations above
Therefore : B₁ / B₂ = 0.84 also r / R = 0.84
Hence ( r ) = 0.84 R
Therefore we can conclude that The point ( r ) within the wire where the field strength equals 84% of the field strength at the wire surface is : 0.84 R
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