The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre

Q: Why did the radius change when the dipole moment doubled?

The dipole moment is the product of current and area ($m = IA$). Since the problem states the current is kept constant, the only way to double the dipole moment is to double the area. Doubling the area means the radius must increase by a factor of $\sqrt{2}$.

Q: What is the formula for the magnetic field at the center of a circular loop?

The magnetic field at the center of a circular loop carrying current $I$ with radius $R$ is given by $B = \frac{\mu_0 I}{2R}$. This is derived from the Biot-Savart Law.

Q: Why didn't we change the number of turns to double the dipole moment?

The problem implies a single circular loop. While increasing the number of turns would indeed increase the dipole moment ($m = NIA$), the standard interpretation of 'a circular loop' without mentioning turns implies $N=1$. Changing the geometry (radius) is the intended path here.

Q: Does the direction of the magnetic field change?

No, as long as the direction of the current remains the same, the direction of the magnetic field at the center will remain the same, as dictated by the Right-Hand Grip Rule.

Q: What if the loop was a square instead of a circle?

The logic would be identical! The area of a square is $a^2$. To double the area, the side length $a$ would become $\sqrt{2}a$. The magnetic field at the center of a square loop is proportional to $I/a$, so the ratio of the fields would still be $\sqrt{2}$.

Question

The dipole moment of a circular loop carrying a current $I$, is $m$ and the magnetic field at the centre of the loop is $B_1$. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $B_2$. The ratio $\frac{B_1}{B_2}$ is

Options:

(A) $2$

(B) $\sqrt{3}$

(C) $\sqrt{2}$

(D) $\frac{1}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a circular loop of wire carrying a steady current $I$. This simple setup is the foundation of electromagnetism.

When current flows in a loop, it generates a magnetic field. At the very center of this loop, the magnetic field $\vec{B}$ points straight along the axis.

Simultaneously, this current loop behaves exactly like a tiny bar magnet. It possesses a **magnetic dipole moment** $\vec{m}$, which is a measure of its magnetic strength and orientation.

### The Master Equations

To solve this problem, we need two fundamental equations from our electromagnetism toolkit.

First, the magnetic dipole moment $m$ of a planar loop is simply the product of the current $I$ and the area $A$ it encloses. For a circular loop of radius $R$, the area is $\pi R^2$.
$$m = IA = I\pi R^2$$

Second, the magnitude of the magnetic field $B$ at the exact center of a circular loop is given by the Biot-Savart Law.
$$B = \frac{\mu_0 I}{2R}$$

Notice a crucial difference here: the dipole moment depends on the **square** of the radius, while the magnetic field depends **inversely** on the radius.

### Finding the New Radius

The problem presents a fascinating scenario: the dipole moment is doubled ($m' = 2m$), but the current $I$ is strictly kept constant.

How can the dipole moment increase if the current doesn't? The only variable left is the area! The loop must have expanded. Let's call the initial radius $R_1$ and the new radius $R_2$.

We can set up an equation based on the doubled dipole moment:
$$I\pi R_2^2 = 2(I\pi R_1^2)$$

Since the current $I$ and $\pi$ are constant, they beautifully cancel out from both sides.
$$R_2^2 = 2R_1^2$$

Taking the square root of both sides reveals the relationship between the new and old radius.
$$R_2 = \sqrt{2}R_1$$

The loop has expanded, and its new radius is $\sqrt{2}$ times larger than the original!

### The Final Ratio

Now that we know the new radius, we can determine the new magnetic field $B_2$ at the center of the expanded loop.

We substitute our expression for $R_2$ into the magnetic field formula.
$$B_2 = \frac{\mu_0 I}{2R_2} = \frac{\mu_0 I}{2(\sqrt{2}R_1)}$$

The question asks for the ratio of the initial magnetic field $B_1$ to the final magnetic field $B_2$. Let's divide them.
$$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R_1}}{\frac{\mu_0 I}{2\sqrt{2}R_1}}$$

Watch how elegantly the physics simplifies. The constants $\mu_0$, $I$, and $2R_1$ cancel out entirely.
$$\frac{B_1}{B_2} = \sqrt{2}$$

The initial magnetic field was stronger by a factor of $\sqrt{2}$. This makes perfect physical sense: as the loop expanded to double its dipole moment, its center moved further away from the current-carrying wire, resulting in a weaker magnetic field at that central point.

Select Answer:

Visualized Solution

$m$ and $B$ of a Loop

Formulas for $m$ and $B$

Initial State

Doubling the Dipole Moment

Finding the New Radius $R_{2}$

Finding the New Magnetic Field $B_{2}$

Calculating the Ratio $\frac{B _{1}}{B _{2}}$

Conclusion

The Sigma Insight: Magnetic Moment of Current Loop

Analyzing the Setup

The Master Equations

Finding the New Radius

The Final Ratio

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