Question

A circular coil of radius \(R\) carries a current \(I\). The magnetic field at its centre is \(B\). At what distance from the centre on the axis of the coil will the magnetic field be \(B/8\)?

• A. \(R\)
• B. \(2R\)
• C. \(\sqrt{3}R\)
• D. \(4R\)

Hint:

Divide the axial-field expression by the field at the centre first. Work with the dimensionless variable $x/R$ so all constants cancel immediately.

Answer 1

The Correct Option is C

Concept: Magnetic field on the axis of a circular current loop at a distance \(x\) from the centre is \[ B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] Magnetic field at the centre of the coil is \[ B_{centre} = \frac{\mu_0 I}{2R} \]

Step 1: Relate the field on axis to the field at the centre. Given \[ B_{axis} = \frac{B_{centre}}{8} \]

Step 2: Substitute the expressions for magnetic fields. \[ \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8}\left(\frac{\mu_0 I}{2R}\right) \] Canceling \( \mu_0 I \) and simplifying, \[ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R} \]

Step 3: Solve for \(x\). \[ (R^2 + x^2)^{3/2} = 8R^3 \] Taking cube root on both sides, \[ R^2 + x^2 = 4R^2 \] \[ x^2 = 3R^2 \] \[ x = \sqrt{3}R \] \[ \boxed{x = \sqrt{3}R} \]

Answer 2

Concept:

• Normalise the axial field by dividing it by the central field.
• This removes $\mu_0$, $I$, and the dimensions of $R$, leaving an equation only in the ratio $x/R$.

Step 1: Write the dimensionless field ratio.
For a circular loop,
$\dfrac{B_x}{B_0}=\left[1+\left(\dfrac{x}{R}\right)^2\right]^{-3/2}$

Step 2: Insert the required reduction.
$B_x=B_0/8$, so
$\left[1+\left(\dfrac{x}{R}\right)^2\right]^{-3/2}=\dfrac18$

Step 3: Remove the power.
Taking both sides to the power $-2/3$ gives
$1+\left(\dfrac{x}{R}\right)^2=8^{2/3}=4$

Step 4: Solve for the distance ratio.
$\left(\dfrac{x}{R}\right)^2=3$
$\dfrac{x}{R}=\sqrt3$, so $x=\sqrt3R$.

Final Answer: $\sqrt3R$, option C.