in a coaxial straight cable the central conductor and the outer conductor carry equal currents in opposite directions magnetic field is zero -outside the cable why outside and not at any other place
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The magnetic field due to the current in central conductor at a point outside the cable, is cancelled by the magnetic field due to the current in the outer conductor.
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Answered by
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see diagram.
We recall the Ampere's circuital law for magnetic field.
Around a cylindrical cable (long one) the magnetic field B is along the tangential direction. This can be known from right hand thumb rule. B will be equal all along the circular path. So integral on LHS = B * 2 π d , d = distance of point from center of cylindrical wire.
On the RHS , i = algebraic sum of all currents enclosed (going out/coming in) with in the circular path.
If there was only one conductor wire: then B * 2 π d = μ₀ I
So we know B.
As there are two conductors, and currents are in opposite direction, RHS = 0. Hence B = 0 outside both conductor wires.
=========================
In between two conductor wires, circular path 2...
RHS = I as we do not include the current of the outer cable.
so B = μ₀ I / (2 π d)
===================
Inside the inner cylindrical wire:
let radius of inner cable = R
let the radius of circular path chosen be = d
amount of current passing inside that circle: I * π d²/ π R²
so B * 2 π d = μ₀ I * d² / R²
We recall the Ampere's circuital law for magnetic field.
Around a cylindrical cable (long one) the magnetic field B is along the tangential direction. This can be known from right hand thumb rule. B will be equal all along the circular path. So integral on LHS = B * 2 π d , d = distance of point from center of cylindrical wire.
On the RHS , i = algebraic sum of all currents enclosed (going out/coming in) with in the circular path.
If there was only one conductor wire: then B * 2 π d = μ₀ I
So we know B.
As there are two conductors, and currents are in opposite direction, RHS = 0. Hence B = 0 outside both conductor wires.
=========================
In between two conductor wires, circular path 2...
RHS = I as we do not include the current of the outer cable.
so B = μ₀ I / (2 π d)
===================
Inside the inner cylindrical wire:
let radius of inner cable = R
let the radius of circular path chosen be = d
amount of current passing inside that circle: I * π d²/ π R²
so B * 2 π d = μ₀ I * d² / R²
Attachments:
in Ampere's law B * dl , the length becomes the circumference along the circular path.... so it is 2 pi R... If you take a rectangular path like it is done in a solenoid, then length of the edge of rectangle is taken...
Alternately, The magnetic field due to the current in central conductor at a point outside the cable, is cancelled by the magnetic field due to the current in the outer conductor. The magnetic field strength is proportional to current divided by distance of the point from the central axis of the cable. So they are equal in magnitude and opposite in sign.
both central and outer conductor forms cable r8?
one cable consists of two conductors. there is an insulation in between the two. Then there is a insulating cover on top of both. The distance for magnetic field "d" is measured from the center of either conductor. So it is same in both.
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Answered by
6
same as the given answer above.
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