Two concentric circular loops, one of radius R and the other of radius 2R, lie in the xy-plane with

Q: Why does the magnetic field have to be perpendicular to the xy-plane?

Because the magnetic field is generated by the cross product of the current element $d\vec{l}$ and the position vector $\vec{r}$. Since both lie in the xy-plane, their cross product is strictly along the z-axis.

Q: How do we know the field depends only on the radial distance $r$?

The system of concentric circular loops is rotationally symmetric about the z-axis. Rotating the coordinate system around the z-axis doesn't change the physical setup, so the field magnitude can only depend on the distance from the origin, $r = \sqrt{x^2 + y^2}$.

Q: Why is the net field at the center pointing inwards?

The outer loop creates an inward field $B_2 = \frac{\mu_0 I_2}{4R}$, and the inner loop creates an outward field $B_1 = \frac{\mu_0 I_1}{2R}$. Since we are given $I_2 > 2I_1$, it mathematically follows that $B_2 > B_1$, making the net field inward.

Q: How can we be sure the field becomes zero somewhere inside the inner loop?

We established the field is inward at the center. However, as you approach the inner wire, the outward field $B_1$ approaches infinity, dominating the finite inward field $B_2$. By the Intermediate Value Theorem, a continuous function transitioning from a negative (inward) to a positive (outward) value must cross zero.

Q: Why is the field between the loops pointing inwards?

In the region $R < r < 2R$, you are outside the inner loop (whose field points inward here) and inside the outer loop (whose field also points inward). The sum of two inward fields is strictly inward.

Question

Two concentric circular loops, one of radius R and the other of radius 2R, lie in the xy-plane with the origin as their common center, as shown in the figure. The smaller loop carries current $I_1$ in the anti-clockwise direction and the larger loop carries current $I_2$ in the clockwise direction, with $I_2 > 2I_1$. $\vec{B}(x, y)$ denotes the magnetic field at a point $(x, y)$ in the xy-plane. Which of the following statement(s) is(are) current?

Options:

(A) $\vec{B}(x, y)$ is perpendicular to the xy-plane at any point in the plane

(B) $|\vec{B}(x,y)|$ depends on x and y only through the radial distance $r = \sqrt{x^2 + y^2}$

(C) $|\vec{B}(x,y)|$ is non-zero at all points for $r < R$

(D) $\vec{B}(x, y)$ points normally outward from the xy-plane for all the points between the two loops

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Concentric Loops Imagine you are looking down at a flat table, which we will call the $xy$-plane. On this table, we have placed two perfectly concentric circular wire loops. The inner loop has a radius $R$ and carries a current $I_1$ flowing in the anti-clockwise direction. Surrounding it is the outer loop, with a larger radius $2R$, carrying a current $I_2$ in the clockwise direction. Before we dive into the physics, there is a crucial mathematical constraint given in the problem: **$I_2 > 2I_1$**. This seemingly simple inequality is the master key that will unlock the behavior of the magnetic field at the center of our system. Keep it in mind as we analyze each option. ### Analyzing Option A: The Direction of the Magnetic Field Let's start by determining the general direction of the magnetic field anywhere in the $xy$-plane. To do this, we invoke the fundamental **Biot-Savart Law**. According to the law, the tiny magnetic field $d\vec{B}$ produced by an infinitesimally small current element $d\vec{l}$ is given by the cross product: $$d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} imes \vec{r}}{r^3}$$ Notice the geometry here. Since both of our wire loops lie entirely flat on the $xy$-plane, any current element $d\vec{l}$ you pick will be a vector in the $xy$-plane. Similarly, if you want to find the magnetic field at any point $(x, y)$ on the table, the position vector $\vec{r}$ from the wire to that point will also lie strictly in the $xy$-plane. What happens when you take the cross product of two vectors that are both in the $xy$-plane? The resulting vector must be perpendicular to both of them, meaning it points straight up or straight down along the $z$-axis. Therefore, the total magnetic field $\vec{B}(x, y)$ is **perpendicular to the $xy$-plane at any point in the plane**. Option A is absolutely correct. ### Analyzing Option B: The Power of Symmetry Physics loves symmetry, and this problem is a beautiful example of it. Look at our system: two perfect, concentric circles. This configuration possesses complete **rotational symmetry** about the $z$-axis. If you were to close your eyes while I rotated the entire setup by $30^\circ$, or $45^\circ$, or any angle, you wouldn't be able to tell the difference when you opened them. Because the physical setup is invariant under rotation, the resulting magnetic field magnitude must also be invariant under rotation. This means the magnitude of the magnetic field, $|\vec{B}(x,y)|$, cannot depend on the specific $x$ or $y$ coordinates individually. It can only depend on how far away you are from the center of symmetry. That distance is the radial distance, $r = \sqrt{x^2 + y^2}$. Thus, Option B is correct. ### Analyzing Option C: The Field Inside the Inner Loop Option C claims that the magnetic field is non-zero everywhere inside the inner loop ($r < R$). To test this, let's act like detectives and check the extreme points of this region. We will start at the very center, where $r = 0$. Using the right-hand thumb rule, the anti-clockwise current $I_1$ in the inner loop produces a magnetic field $B_1$ pointing **outwards** ($\odot$) from the page. The clockwise current $I_2$ in the outer loop produces a magnetic field $B_2$ pointing **inwards** ($\otimes$) into the page. Their magnitudes are: $$B_1 = \frac{\mu_0 I_1}{2R} \quad ext{and} \quad B_2 = \frac{\mu_0 I_2}{4R}$$ Now, remember our master key? We are given that $I_2 > 2I_1$. Let's substitute this into our expression for $B_2$: $$B_2 = \frac{\mu_0 I_2}{4R} > \frac{\mu_0 (2I_1)}{4R} = \frac{\mu_0 I_1}{2R} = B_1$$ This proves that $B_2 > B_1$. The inward field from the outer loop is stronger than the outward field from the inner loop at the center. Therefore, the **net magnetic field at the origin points inwards**. But what happens as we move away from the center and get extremely close to the inner wire? As the distance to a wire approaches zero, the magnetic field it produces approaches infinity. So, just inside the inner loop (as $r o R$), the outward field $B_1$ becomes infinitely large, completely overpowering the finite inward field $B_2$. Think about the profound implication of this: the net magnetic field is pointing **inwards** at the center, but it is pointing **outwards** near the edge of the inner loop. For a continuous physical field to flip its direction from negative (inward) to positive (outward), it must cross zero somewhere in between! Therefore, there must be at least one point in the region $r < R$ where the magnetic field is exactly zero. Option C is incorrect. ### Analyzing Option D: The Annular Region Between the Loops Finally, let's examine the region between the two loops, where $R < r < 2R$. In this annular space, you are standing outside the inner loop but inside the outer loop. Let's apply the right-hand rule one last time. For a point outside the anti-clockwise inner loop, its magnetic field $B_1$ points **inwards** ($\otimes$). For a point inside the clockwise outer loop, its magnetic field $B_2$ also points **inwards** ($\otimes$). Since both individual magnetic fields are pointing into the plane, their vector sum must also point normally **inwards**. Option D claims the field points normally outward, which is the exact opposite of reality. Option D is incorrect. ### Conclusion By systematically applying the Biot-Savart law, leveraging rotational symmetry, and carefully analyzing the boundary conditions using the given current constraints, we have successfully decoded the magnetic field's behavior. The only correct statements are A and B.

Select Answer:

Visualized Solution

$System Analysis$

$Direction of B$

$Symmetry of the Field$

$Field at the Center (r = 0)$

$Net Field at Center$

$Field near the Inner Loop (r \to R)$

$Field Between the Loops (R < r < 2 R)$

The Sigma Insight: Biot-Savart Law

The Setup

Analyzing Option A

Analyzing Option B

Analyzing Option C

Analyzing Option D

Conclusion

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction of B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Symmetry of the Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field at the Center (r=0)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Field at Center

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field near the Inner Loop (r→R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field Between the Loops (R<r<2R)