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If the current density in a linear condu...

If the current density in a linear conductor of radius 'a' varies with r according to relation `J=kr^2`, where k is a constant and r is the distance of a point from the axis of conductor. Find the magnetic field induction at a point distance r from the axis, when (i) `rlta` and (ii) `rgta`.

Text Solution

Verified by Experts

The correct Answer is:
(i) `(mu_0mu_rkr^3)/(r)` (ii) `(mu_0ka^4)/(4r)`
(i) Consider an elementary ring of radius r thickness dr whose centre lies on the axis of the conductor. Area of the elementary ring, `dA=2pirdr`. Current passing through the elementary ring is
`dI=JdA=(kr^2)(2pirdr)=k2pir^3dr`
Total current passing through the closed path of radius r is,
`I=int_(0)^(r)k2pir^3dr=(kpir^4)/(2)`
Using Ampere circuital law, we have
`oint vecB.dvecl=mu_0mu_rI`
`B2pir=mu_0mu_r(pikr^4)/(2)` or `B=(mu_0mu_rkr^3)/(4)`
(ii) If `r gt a`, then net current through the closed path of radius r is,
`I=int_(0)^(a) k2pir^3dr=(pika^4)/(2)`
Using Ampere circuital law, we have
`oint vecB.dvecl=mu_0I` or `B2pir=(mu_0pika^4)/(2)`
`B=(mu_0ka^4)/(4r)`
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Knowledge Check

  • Ampere's law provides us an easy way to calculate the magnetic field due to a symmetrical distribution of current. Its mathemfield expression is ointvecB.dl=mu_0I_("in") . The quantity on the left hand side is known as line as integral of magnetic field over a closed Ampere's loop. If the current density in a linear conductor of radius a varies with r according to relation J=kr^2 , where k is a constant and r is the distance of a point from the axis of conductor, find the magnetic field induction at a point distance r from the axis when rlta. Assume relative permeability of the conductor to be unity.

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