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Question
The figure shows a cylindrical wire of diameter a, carrying a current I. The current density in the wire is in the direction of the axis and varies linearly with radial distance r from the axis according to the relation J = `"J"_0 "r"/"a"`. Obtain a magnetic field B inside the wire at a distance r from its center.
Solution
dI = JdA = `("J"_0"r"/"a") = 2 pi "rdr" = (2pi"J"_0"r"^2"dr")/"a"` .....(1)
When we apply Ampere's circuital law to the circular path of integration, we see that the wire has perfect cylindrical symmetry, with all charges moving parallel to the wire. As a result, the magnetic field must be tangent to circles concentric with the wire. The enclosed current is the current that is enclosed within radius r. Thus,
`oint "Bdl" = mu_0"I"_"encl"`
`therefore oint "Bdl" = mu_0 int_0^"r" "dl" = mu_0 int_0^"r" (2pi"J"_0)/"a" "r"^2 "dr"` ....(2)
`therefore "B"(2pi"r") = (mu_0 2pi"J"_0)/"a" ("r"^3/3)`
`therefore "B" = (mu_0 "J"_0)/"3a" "r"^2` ....(3)
which is the required expression.
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