Discussion Forum

 The magnitude of the magnetic induction  at a point on the axis  at a  large distance (r) from  the centre of a circular coil of  'n' turns and area 'A' carrying current (l)  is given by

Answer:

Explanation:

As we know that the magnetic field on the axis of a circular current carrying loop

$B_{axis}=\frac{\mu_{0}nla^{2}}{2(r^{2}+a^{2})^{3/2}}$..............(i)

where, l= current tjhrough 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= $\pi  a^{2} $

 $\Rightarrow$     $ a^{2}=\frac{A}{\pi}$   ..........(ii)

and it r>>a then,  $(r^{2}+a^{2})^{3/2}=r^{3}$   ............(iii)

From Eqs.(i) , (ii) and (iii) we get

$B_{}=(\frac{\mu_{0}nl}{2r^{3}})\frac{A}{\pi}\times\frac{2}{2}$

   $\Rightarrow$      $ B_{}=\frac{2\mu_{0}nlA}{4\pi r^{3}}$

So, option (b) is correct