Question

A current carrying circular coil of radius 'R' produces magnetic field \(B_1\) at an axial point P at a distance 'x' from its centre and \(B_2\) at point Q placed at its centre respectively . If \(B_2 = 8B_1\), the value of x is

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

Hint:

Compare the axial field formula with the centre value and solve for x.

Answer 1

The Correct Option is B

Step 1: Write the ratio
$\dfrac{B_1}{B_2} = \dfrac{R^3}{(R^2 + x^2)^{3/2}} = \dfrac{1}{8}$.

Step 2: Take the cube root
$\dfrac{R}{(R^2 + x^2)^{1/2}} = \dfrac{1}{2}$, so $\sqrt{R^2 + x^2} = 2R$.

Step 3: Square
$R^2 + x^2 = 4R^2$ gives $x = \sqrt{3}R$.

Step 4: Check
The ratio at $x = \sqrt{3}R$ is $\left(\dfrac{R}{2R}\right)^3 = \dfrac{1}{8}$, as given.

Final Answer:
The distance is sqrt 3 R. This is option (B). \[ \boxed{\text{(B) }\sqrt{3}R} \]