(i) The electric current passing through the wire is the conduction current `I_(C )`. This current generates magnetic field around the wire connected across the capacitor.
(ii) Therefore , when a magnetic needle is kept near the wire, deflection is observed . in order to compute the strength of magnetic field at a point, we use Ampere's circuital law is used which states that ' the line
integral of the magnetic field `vec(B)` around any closed loop is equal to `mu_(0)` times the net current I threading through the area enclosed by the loop'. Ampere's law in equation form is
`oint vec(B).d vec(l) = mu_(0)`I(t)
(iii) where `mu_(0)` is the permeability of free space. To calculate the magnetic field at a point P near the wire as shown in Figure, let us drow an amperean loop is drawn (circular loop ) which encloses the surface `S_(1)` (circular surface ). Therefore, using Ampere's circuital law (equation 1 ),
`underset(s_(1))(int) vec(B). d vec(l) = mu_(0)I_(C )`
where `I_(c)` is the conduction current.
(iv) Suppose the same loops is enclosed by balloon shaped surface `S_(2)`. this means that the boundaries of two surface `S_(1)` and `S_(2)`a re same but shape of the enclosing surfaces are different (first surface `(S_(1))` is circular in shape and second one is balloon shaped surface `(S_(2))`. as the Ampere's law applied for a given closed loop does not depend on shape of the enclosing surface, the integrals will give the same answer. but by appling Ampere's circuital law (equation 1),
(v) The right hand side of equation is zero because the surface `S_(2)` no where touches the wire carrying coduction current and further, there is no current in between the plates of the capacitor (there is a discontinuity). so the magnetic field at a point P is zero. Hence there is an inconsistency between equation (5.4) and equation (2). J. C.Maxwell resolved this inconsistency as follows:
(vi) Due to external source (battery or cell), the capacitor gets charged up because of current flowing through the capacitor. this produces an increasing electric field between the capacitor plates. so, there must be a current associated with the changing electric field in between the capacitor plates. in other words, the time verying electric flux (or time varying electric field existing between the plates of the capacitor aslo produces a current known as displacement current.
(vii) From Gauss's law (refer Unit 1), the electric flux between the plates of the capacitor (figure is)
`Phi_(E ) =oint oint vec(E).d vec(A) = EA = (q)/(E_(0))`
where A is the area of the plates of capacitor the change in electric flux is
`(d Phi_(E))/(dt) = (1)/(epsilon_(0)) (dq)/(dt) rArr (dq)/(dt) = I_(d) = epsilon (d Phi_(E))/(dt)`
where, `I_(d)` is known as displacement current.
(viii) The displacement current can be defined as the current which comes into play in the region in which the electric field and the electric flux are changing with time. In other words, whenever the change in electric field takes place, displacement current is produced. Maxwell modified Ampere's law as
`oint vec(B).d vec(S) = mu_(0) (I_(c ) + I_(d))`
where I + `I_(c ) + I_(d)` which means the total current enclosed by the surface is sum of conduction current and displacement current.
(ix) When a constant current is applied, displacement current `I_(d)` = 0 and hence `I_(c)` = I. Between the plates, the conduction current `I_(c )` = 0 hence `I_(d)` = I.
