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 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:
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For \( x = \sqrt{3}r \), the axial field is exactly one-eighth of the field at the center of the coil.
Updated On: Jun 9, 2026
- \( \frac{1}{8} \text{T} \)
- \( \frac{1}{16} \text{T} \)
- \( \frac{1}{4} \text{T} \)
- \( \frac{1}{12} \text{T} \)
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The Correct Option is A
Solution and Explanation
Concept:
The magnetic field \( B_c \) at the center of a circular coil of radius \( r \) carrying current \( I \) is \( B_c = \frac{\mu_0 I}{2r} \). The magnetic field \( B_a \) at a distance \( x \) on the axis from the center is given by \( B_a = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \).
Step 1: Relate axial field to center field.
The formula for the field on the axis can be expressed as a ratio to the center field: $$ B_a = B_c \cdot \frac{r^3}{(r^2 + x^2)^{3/2}} $$
Step 2: Substitute the given parameters.
Given \( x = \sqrt{3}r \) and \( B_c = 1 \text{ T} \): $$ B_a = 1 \cdot \frac{r^3}{(r^2 + (\sqrt{3}r)^2)^{3/2}} $$ $$ B_a = \frac{r^3}{(r^2 + 3r^2)^{3/2}} = \frac{r^3}{(4r^2)^{3/2}} $$
Step 3: Simplify the expression.
$$ B_a = \frac{r^3}{(4^{3/2} \cdot r^{3/2 \times 2})} = \frac{r^3}{8r^3} = \frac{1}{8} \text{ T} $$ $$\boxed{\frac{1}{8} \text{ T}}$$
Step 1: Relate axial field to center field.
The formula for the field on the axis can be expressed as a ratio to the center field: $$ B_a = B_c \cdot \frac{r^3}{(r^2 + x^2)^{3/2}} $$
Step 2: Substitute the given parameters.
Given \( x = \sqrt{3}r \) and \( B_c = 1 \text{ T} \): $$ B_a = 1 \cdot \frac{r^3}{(r^2 + (\sqrt{3}r)^2)^{3/2}} $$ $$ B_a = \frac{r^3}{(r^2 + 3r^2)^{3/2}} = \frac{r^3}{(4r^2)^{3/2}} $$
Step 3: Simplify the expression.
$$ B_a = \frac{r^3}{(4^{3/2} \cdot r^{3/2 \times 2})} = \frac{r^3}{8r^3} = \frac{1}{8} \text{ T} $$ $$\boxed{\frac{1}{8} \text{ T}}$$
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