Question

The magnetic field at the centre of a circular current carrying-conductor of radius $r$ is $B_c$ . The magnetic field on its axis at a distance r from the centre is $B_a$ . The value of $B_c:B_a$ will be :

• A. $1:\sqrt{2}$
• B. $1:2\sqrt{2}$
• C. $2\sqrt{2}:1$
• D. $\sqrt{2}:1$

Answer 1

Answer: (C)

Magnetic field at centre of current carrying coil,
$B_c=\frac{{\left(\mu \right)}_{0}I}{2r}...(i)$
Magnetic field at axial point due to a currentcarrying coil at distance of $r$ ,
$d=r$
$B=\frac{{\left(\mu \right)}_{0}I{r}^2}{2{\left(r^2+d^2\right)}^{3/2}}$
$\Rightarrow B_a=\frac{{\left(\mu \right)}_{0}I{r}^2}{2{\left(2r^2\right)}^{3/2}}....(ii)$
$Now,\frac{B_c}{B_a}=\frac{{\left(\mu \right)}_{0}I}{2r}\times \frac{2{\left(2r^2\right)}^{3/2}}{{\left(\mu \right)}_0{\left(Ir\right)}^2}=2\sqrt{2}$
$B_c:B_a=2\sqrt{2}:1$