Question

The magnetic field at the center of a circular current-carrying loop of radius $R$ is $B_0$. The magnetic field at an axial point at a distance $x = R$ from the center of the loop is:

• A. \( B_0/2 \)
• B. \( B_0/2\sqrt{2} \)
• C. \( B_0/4 \)
• D. \( B_0\sqrt{2} \)

Hint:

For rapid calculations on a circular loop's axis, remember the ratio formula: \( \frac{B_{\text{axial}}}{B_{\text{center}}} = \left(1 + \frac{x^2}{R^2}\right)^{-3/2} \). Substituting \(x = R\) directly yields \((1+1)^{-3/2} = 2^{-3/2} = \frac{1}{2\sqrt{2}}\).

Answer 1

The Correct Option is B

Topic of the Question:
This question deals with electromagnetism, specifically the calculation and comparison of the magnetic field generated by a circular current-carrying loop at its center versus along its axis.
Step 1 : Understanding the Question:
We are given that a circular loop of radius $R$ carries a current and produces a magnetic field of magnitude $B_0$ at its center. We need to determine the magnitude of the magnetic field at a point on the axis of the loop at a distance $x = R$ from the center.
Step 2 : Key Formulas and Approach:

The magnetic field at the center of a circular loop carrying current $I$ is: $B_{\text{center}} = \frac{\mu_0 I}{2R}$.

The magnetic field at an axial point located at a distance $x$ from the center of the loop is: $B_{\text{axial}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.

We can solve this by expressing the axial magnetic field formula in terms of the central magnetic field $B_0$ and substituting $x = R$.

Step 3 : Detailed Explanation:

We begin by writing the expression for the magnetic field at the center of the loop, which is given as $B_0$: $B_0 = \frac{\mu_0 I}{2R}$.

Next, we write down the general equation for the magnetic field along the axis of the loop: $B_{\text{axial}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.

We substitute the given axial distance $x = R$ into this equation: $B_{\text{axial}} = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}}$.

Simplifying the term in the denominator: $(R^2 + R^2)^{3/2} = (2R^2)^{3/2} = 2^{3/2} \cdot (R^2)^{3/2} = 2\sqrt{2} R^3$.

We substitute this back into our expression for the axial field: $B_{\text{axial}} = \frac{\mu_0 I R^2}{2 \cdot 2\sqrt{2} R^3} = \frac{\mu_0 I}{2R \cdot 2\sqrt{2}}$.

By comparing this simplified expression to our initial equation for $B_0$, we can substitute $B_0 = \frac{\mu_0 I}{2R}$ directly into the relation.

This substitution yields the final axial field expression: $B_{\text{axial}} = \frac{B_0}{2\sqrt{2}}$.

Step 4 : Final Answer:
The magnetic field at the axial distance $x = R$ is equal to $B_0 / (2\sqrt{2})$, which matches Option (B).