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.
In the above question, find the magnetic field induction at a point distance r from the axis when rgta. Assume relative permeability of the medium surrounding the conductor to be unity.

Options

A. `(mu_0ka^4)/(4r)`
B. `(mu_0kr^3)/2`
C. `(mu_0kpia^4)/(2r)`
D. `(mu_0ka^3)/4`

Answer

A

Explanation

(a)
Magnetic field at any point on Ampere's loop can be due to
all currents passing through inside or outside the loop. But net
contribution in the left hand side will come from inside current only.
For rlta, current passing through within the cylinder of radius
r is given by
`int_0^rJdA=int_0^rkr^2 2pirdr=2pikint_0^r r^3dr`
`=kpir^4//2`
Now using Ampere's law:
`Bxx2pir=mu_0I=mu_0kpir^4//2 implies B=(mu_0kr^3)/4`
For `rgta, I=kpia^4//2` (by putting r=a)
`:. B=(mu_0ka^4)/(4r)`