Answer:
Explanation:
We shall have to apply Ampere's circuital law to find out the magnetic field asked for.
Ampere's circuital law expression is given below :
∫ B dl = μ₀ I
B is magnetic field , I is current inside the closed curve over which line integral of magnetic field is taken
If we take a point at distance r from the axis of wire , Th magnetic field B at that point can be calculated as follows
current enclosed by the closed circle of radius r
I = cross sectional area x current density
π ( R₂² - R₁² ) x J
Applying Ampere's law
∫ B dl = μ₀ I
B x 2π r = μ₀ x π ( R₂² - R₁² ) x J
B = μ₀ x ( R₂² - R₁² ) x J / 2 r
[tex]=\frac{\mu_0J(R_2^2-R_1^2)}{2r}[/tex].
The magnitude of the magnetic field is [tex]B = \frac{\mu_o J(R_2^2 - R_1^2)}{2r}[/tex].
The given parameters;'
Apply Ampere's law to determine the magnitude of the magnetic field is calculated as follows;
[tex]\int\limits {B} \, dl = \mu_o I\\\\[/tex]
where;
[tex]B(2\pi r) = \mu_o (JA)\\\\B(2\pi r) = \mu_o J \pi(R_2^2 - R_1^2)\\\\B = \frac{\mu_o J(R_2^2 - R_1^2)}{2r}[/tex]
Thus, the magnitude of the magnetic field is [tex]B = \frac{\mu_o J(R_2^2 - R_1^2)}{2r}[/tex].
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