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 \(B_{2}=27B_{1}\), the value of 'x' is

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

Hint:

Take the ratio of the centre field to the axial field and set it equal to 27.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
At the centre \(B_2=\dfrac{\mu_0I}{2R}\). On the axis at distance \(x\), \(B_1=\dfrac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}\).

Step 2: Take the ratio
\(\dfrac{B_2}{B_1}=\dfrac{(R^2+x^2)^{3/2}}{R^3}=27\)

Step 3: Solve
\(\sqrt{R^2+x^2}=3R\Rightarrow x^2=8R^2\)

Step 4: Result
\(x=2\sqrt{2}\,R\)

Final Answer:
The distance is \(x=2\sqrt{2}\,R\). \[ \boxed{\text{(C) }2\sqrt{2}\text{R}} \]