Question

A circular coil has radius ' \(R\) '. The distance from centre on axis where induction is \(1/27\) th of its value at centre is.

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

Hint:

Magnetic field on axis decreases rapidly as distance from the center increases.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The magnetic field of a circular current-carrying loop decreases as we move along its axis. We need to find the distance \(x\) where the axial field is \(1/27\) of the central field.
Step 2: Key Formula or Approach:
Magnetic field at center: \(B_{C} = \frac{\mu_{0} I}{2R}\).
Magnetic field on the axis at distance \(x\): \(B_{A} = \frac{\mu_{0} I R^{2}}{2(R^{2} + x^{2})^{3/2}}\).
The ratio is \(\frac{B_{A}}{B_{C}} = \left( \frac{R^{2}}{R^{2} + x^{2}} \right)^{3/2}\).
Step 3: Detailed Explanation:
Given \(\frac{B_{A}}{B_{C}} = \frac{1}{27}\).
\[ \frac{1}{27} = \left( \frac{R^{2}}{R^{2} + x^{2}} \right)^{3/2} \]
Take the cube root of both sides:
\[ \left(\frac{1}{27}\right)^{1/3} = \left[ \left( \frac{R^{2}}{R^{2} + x^{2}} \right)^{3/2} \right]^{1/3} \]
\[ \frac{1}{3} = \left( \frac{R^{2}}{R^{2} + x^{2}} \right)^{1/2} \]
Square both sides:
\[ \frac{1}{9} = \frac{R^{2}}{R^{2} + x^{2}} \]
\[ R^{2} + x^{2} = 9R^{2} \]
\[ x^{2} = 8R^{2} \]
\[ x = \sqrt{8} R = 2\sqrt{2} R \]
Step 4: Final Answer:
The distance is \(2\sqrt{2} R\).