An extremely long straight wire, with radius of cross-section a, carries current I. Then ratio of magnetic fields at distances `a/2` and 2a from its axis would be ______ .

Explanation

1. Consider a solid wire with radius of cross-section equal to a. When I amount of current passing through it is distributed uniformly over its cross-section, current density is found to be constant at all the points in the wire.

2. Here, current passing through the wire produces magnetic field at all the points around the axis of wire. Consider following two cases.
3. Inside the wire `(r_(i)lta)` :
(i) Consider Amperean loop 1 which is a circle of radius `r_(i)lta` and centre lying on the axis of current carrying straight wire. (Here, suffix i means inside the wire and so `r_(i)lta`). Plane of this loop is perpendicular to the axis of wire.
(ii) All the points on this loop are symmetric about centre `C_(1)`, and so magnitude of magnetic field at all these points is same.
Let it be `B_(i)`. Direction of `vecB_(i)` at a given point on this circle, is in the direction of tangent drawn to this circle at that point as shown in the figure. If `I_(e)` is the amount of current enclosed by this loop then, since current density is same at all the points on the entire cross-section of the wire,
`I_(e)/(pir_(i)^(2))=I/(pia^(2))`
`thereforeI_(e)=I(r_(i)^(2)/a^(2))" "...(1)`
(iii) Now, applying Ampere.s circuital law to Amperean loop 1 we get,
`intvecB_(i)*vec(dl)=mu_(0)I_(e)`
`thereforeintB_(i)dlcos0^(@)=mu_(0)I_(e)" "(becausevecB_(i)"||"vec(dl))`
`thereforeB_(i)intdl=mu_(0)I_(e)" "(becauseB_(i)="constant")`
`thereforeB_(i)(2pir_(i))=mu_(0)(Ir_(i)^(2)/a^(2))`
`thereforeB_(i)=(mu_(0)I)/(2pia^(2))(r_(i))" "...(2)`
`thereforeB_(i)propr_(i)" "("For "r_(i)lta)" "...(3)`
4. Outside the wire `(r_(0)gta)` :
(i) Consider Amperean loop 2 which is a circle of radius `r_(0)gta` and centre lying on the axis of current carrying straight wire. (Here, suffix O means outside the wire and so `r_(0)gta`). Plane of this loop is perpendicular to the axis of wire.
(ii) All the points on this loop are symmetric about centre `C_(2)` and so magnitude of magnetic field at all these points is same.
Let it be `B_(0)`. Direction of `vecB_(0)` point on this circle is in the direction of S tangent drawn to this circle at that point as shown in the figure. Here, amount of She current enclosed by this loop is `I_(e)=I` = current passing through given wire.
(iii) Now, applying Amperes circuital law to Amperian loop 2, we get,
`intvecB_(0)*vec(dl)=mu_(0)I_(e)`
`thereforeintB_(0)dlcos0^(@)=mu_(0)I" "(because"Here",I_(e)=I)`
`(becausevecB_(0)"||"vec(dl))`
`thereforeB_(0)intdl=mu_(0)I" "(becauseB_(0)="constant")`
`thereforeB_(0)(2pir_(0))=mu_(0)I`
`thereforeB_(0)=(mu_(0)I)/(2pir_(0))" "...(4)`
`thereforeB_(0)prop1/r_(0)" "("For "r_(0)gta)" "...(5)`
(iii) Equations (2) and (4) give required results.
(6) Note : (i) To find magnetic field `B_s` on the surface of currect carrying staright wirre we can write `r_i = a` in equation (2) or `r_0= a` in eqution (4) to get ,
`B_s = (mu_0 I)/(2pia)` ...(6)
(ii) Form results (3), (5) and (6), graph of `B to r` is obtained as follow :