Question

The magnetic field at the centre of a circular coil of radius R carrying current I is 64 times the magnetic field at a distance x on its axis from the centre of the coil. Then the value of x is

• A. \(\frac{R}{4}\sqrt{3}\)
• B. \(R\sqrt{3}\)
• C. \(\frac{R}{4}\)
• D. \(R\sqrt{15}\)

Answer 1

The Correct Option is D

Given:

• Magnetic field at center of circular coil is 64 times the field at a point on its axis at distance \( x \) from center
• Need to find \( x \) in terms of radius \( R \)

Step 1: Magnetic Field at Center of Coil

\[ B_{\text{center}} = \frac{\mu_0 I}{2R} \]

Step 2: Magnetic Field on Axis of Coil

The magnetic field at a point on the axis of a current-carrying circular coil is:

\[ B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

Step 3: Use Given Condition

We are told:

\[ B_{\text{center}} = 64 \cdot B_x \] Substituting both expressions: \[ \frac{\mu_0 I}{2R} = 64 \cdot \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

Step 4: Simplify the Equation

Cancel \( \frac{\mu_0 I}{2} \) on both sides:

\[ \frac{1}{R} = 64 \cdot \frac{R^2}{(R^2 + x^2)^{3/2}} \Rightarrow \frac{1}{R} = \frac{64R^2}{(R^2 + x^2)^{3/2}} \]

Step 5: Cross Multiply and Solve

\[ (R^2 + x^2)^{3/2} = 64R^3 \Rightarrow R^2 + x^2 = (64R^3)^{2/3} = 16R^2 \Rightarrow x^2 = 15R^2 \Rightarrow x = R\sqrt{15} \]

The value of \( x \) is \( \boxed{R\sqrt{15}} \), so the correct answer is (D).