A coil consisting of 100 circular loops with radius $0.60\text{ m}$ carries a current of $5\text{ A}$. At what distance from the center, along the axis, the magnetic field magnitude 1/8 is great as it is at the center?

🧲 Physics Problem: Magnetic Field of a Coil

Question

A coil consisting of $\mathbf{100}$ circular loops with radius $\mathbf{0.60\text{ m}}$ carries a current of $\mathbf{5\text{ A}}$. At what distance from the center, along the axis, the magnetic field magnitude is $\mathbf{1/8}$ as great as it is at the center?

Given Data

• Number of loops, $N = 100$
• Radius of the coil, $a = 0.60\text{ m}$
• Current in the coil, $I = 5\text{ A}$

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⚙️ Step-by-Step Solution

[Image of Magnetic field lines for a circular current loop]

1. Formulas for Magnetic Field

The magnetic field ($B_x$) along the axis at distance $x$ and the magnetic field at the center ($B_{\text{center}}$) are:

$$B_x = \frac{\mu_0 N I a^2}{2(a^2 + x^2)^{3/2}}$$

$$B_{\text{center}} = \frac{\mu_0 N I}{2a}$$

2. Setting up the Condition

The problem states that the field at distance $x$ is $\mathbf{1/8}$ of the field at the center: $\mathbf{B_x = \frac{1}{8} B_{\text{center}}}$.

$$\frac{\mu_0 N I a^2}{2(a^2 + x^2)^{3/2}} = \frac{1}{8} \left( \frac{\mu_0 N I}{2a} \right)$$

3. Simplification and Solving for $x$

We cancel the common terms ($\mu_0$, $N$, $I$, and $2$) from both sides and cross-multiply:

$$\frac{a^2}{(a^2 + x^2)^{3/2}} = \frac{1}{8a} \quad \Rightarrow \quad 8a^3 = (a^2 + x^2)^{3/2}$$

Raise both sides to the power of $\mathbf{2/3}$ to simplify the exponent:

$$(8a^3)^{2/3} = a^2 + x^2 \quad \Rightarrow \quad 4a^2 = a^2 + x^2$$

Isolating $x^2$ and taking the square root gives the symbolic answer:

$$x^2 = 3a^2 \quad \Rightarrow \quad \mathbf{x = \sqrt{3} a}$$

4. Final Calculation

Substituting the radius $a = 0.60\text{ m}$:

$$x = \sqrt{3} \cdot (0.60\text{ m}) \approx 1.0392\text{ m}$$

The required distance is approximately $\mathbf{1.04\text{ m}}$.