Question

The magnetic induction due to circular current-carrying conductor of radius $a$, at its centre is $B_c$. The magnetic induction on its axis at a distance $a$ from its centre is $B_a$. The value of $B_c:B_a$ will be :

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

Answer 1

Answer: (C)

$2\sqrt{2}:1$

The magnetic induction due to circular current-carrying conductor of radius $a$, at its centre is
$B_c=\frac{{\mu }_{0}I}{2a}$ (1)

The magnetic field on the axis of a current-carrying circular coil of radius $r$ at a distance $d$ from its center is given by
$B=\frac{{\mu }_{0}I{r}^2}{2(d^2+r^2{)}^{\frac{3}{2}}}$

Here, $r=d=a$, hence

$B_a=\frac{{\mu }_{0}I{a}^2}{2(a^2+a^2{)}^{\frac{3}{2}}}$

or, $B_a=\frac{{\mu }_{0}I{a}^2}{2(2a^2{)}^{\frac{3}{2}}}$

or, $B_a=\frac{{\mu }_{0}I}{4\sqrt{2}a}$ (2)

Therefore, from (1) and (2),

$\frac{B_c}{B_a}=2\sqrt{2}$