Question

A current carrying circular coil of radius 'r' produces a magnetic induction of 1 T at its centre. The magnetic induction at a distance of \(\sqrt{3}r\) on its axis from its centre is:

• A. \( \frac{1}{8} \text{T} \)
• B. \( \frac{1}{16} \text{T} \)
• C. \( \frac{1}{4} \text{T} \)
• D. \( \frac{1}{12} \text{T} \)

Hint:

For \( x = \sqrt{3}r \), the axial field is exactly one-eighth of the field at the center of the coil.

Answer 1

The Correct Option is A

Step 1: Centre and axial fields.
A circular coil gives $B_c = \dfrac{\mu_0 I}{2r}$ at its centre and $B_a = \dfrac{\mu_0 I r^2}{2(r^2+x^2)^{3/2}}$ at axial distance $x$.
Step 2: Take the ratio.
Dividing the axial field by the centre field cancels $\mu_0 I$: \[ \frac{B_a}{B_c} = \frac{r^3}{(r^2+x^2)^{3/2}}. \]
Step 3: Substitute $x = \sqrt{3}\,r$.
Then $r^2 + x^2 = r^2 + 3r^2 = 4r^2$.
Step 4: Plug into the ratio.
\[ \frac{B_a}{B_c} = \frac{r^3}{(4r^2)^{3/2}}. \]
Step 5: Simplify the denominator.
$(4r^2)^{3/2} = 4^{3/2}\,r^{3} = 8r^3$, so $\dfrac{B_a}{B_c} = \dfrac{r^3}{8r^3} = \dfrac{1}{8}$.
Step 6: Use $B_c = 1\,\text{T}$.
$B_a = \dfrac{1}{8}\times 1 = \dfrac{1}{8}\,\text{T}$.
\[ \boxed{\dfrac{1}{8}\ \text{T}} \]