A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not constant across

Question

A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not constant across the cross section of the wire, but rather varies as J=αrJ=αr, where αα is a constant.
(a) By the requirement that J integrated over the cross section of the wire gives the total current I, calculate the constant αα in terms of I and R.
(b) Use Ampere’s law to calculate the magnetic field B(r) for (i) r≤Rr≤R and (ii) r≥Rr≥R. Express your answers in terms of I.

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Huyền Thanh 1 year 2021-08-18T17:53:33+00:00 1 Answers 679 views 0

Answers ( )

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    2021-08-18T17:55:10+00:00

    Answer: (a) α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

    (b) For r≤R: B(r) = μ_0.[tex](\frac{I.r^{2}}{2.\pi.R^{3}})[/tex]

    For r≥R: B(r) = μ_0.[tex](\frac{I}{2.\pi.r})[/tex]

    Explanation:

    (a) The current I enclosed in a straight wire with current density not constant is calculated by:

    [tex]I_{c} = \int {J} \, dA[/tex]

    where:

    dA is the cross section.

    In this case, a circular cross section of radius R, so it translates as:

    [tex]I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr[/tex]

    [tex]I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr[/tex]

    [tex]I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}[/tex]

    [tex]\alpha = \frac{3I}{2.\pi.R^{3}}[/tex]

    For these circunstances, α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

    (b) Ampere’s Law to calculate magnetic field B is given by:

    [tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

    (i) First, first find [tex]I_{c}[/tex] for r ≤ R:

    [tex]I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr[/tex]

    [tex]I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr[/tex]

    [tex]I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr[/tex]

    [tex]I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}[/tex]

    [tex]I_{c} = \frac{I.r^{3}}{R^{3}}[/tex]

    Calculating B(r), using Ampere’s Law:

    [tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

    [tex]B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )[/tex].μ_0

    B(r) = [tex](\frac{Ir^{3}}{R^{3}2.\pi.r})[/tex].μ_0

    B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

    For r ≤ R, magnetic field is B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

    (ii) For r ≥ R:

    [tex]I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr[/tex]

    So, as calculated before:

    [tex]I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}[/tex]

    [tex]I_{c} =[/tex] I

    Using Ampere:

    B.2.π.r = μ_0.I

    B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0

    For r ≥ R, magnetic field is; B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0.

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