Find the magnetic field induction at a point on the axis of a circular coil carrying current and hence find the magnetic field at the centre of circular coil carrying current.

Options

A. `B_("axis")=(mu_(0))/(4pi)dot(nA)/(Ir^(3))`
B. `B_("axis")=(mu_(0))/(4pi)dot(2nIA)/(r^(3))`
C. `B_("axis")=(mu_(0))/(4pi)dot(2nI)/(Ar^(3))`
D. `B_("axis")=(mu_(0))/(4pi)dot(nIA)/(r^(3))`

Answer

B

Explanation

As we know that the magnetic field on the axis of a circular current carrying loop,
`B=(mu_(0)nla^(2))/(2(r^(2)+a^(2))^(3//2))` ….(i)
where , I = current through the coil , a = radius of a circular loop, r = distance of point from the centre along the axis and n = number of turns in the coil.
Area of the coil , A= `pia^(2)`
`!a^(2)=A/pi` ...( ii)
and it `rgtgta` then, `(r^(2)+a^(2))^(3//2)~~r^(3)` ...(iii)
From Eqs. (i), (ii) and (iii), we get
`B=((mu_(0)nl)/(2r^(3)))A/pixx2/2`
`!B=(2mu_(0)nlA)/(4pir^(3))`
So, option (b) is correct.