A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I i

Q: Why does the wire specifically take the shape of a circle?

The magnetic force $I(d\vec{l} \times \vec{B})$ acts perpendicular to every segment of the wire, pushing it outwards uniformly. A flexible wire under uniform outward pressure will expand to maximize its enclosed area for a fixed perimeter, which geometrically results in a perfect circle.

Q: Why do we take a small element subtending $2d\theta$ instead of just $d\theta$?

Taking $2d\theta$ is a mathematical convenience. It makes the geometry symmetrical around the vertical axis, so the angle on each side is exactly $d\theta$. This avoids dealing with fractions like $d\theta/2$ when resolving the tension vectors.

Q: Why can we assume $\sin(d\theta) \approx d\theta$?

In calculus, when an angle is infinitesimally small and measured in radians, the value of its sine is approximately equal to the angle itself. This is a standard small-angle approximation derived from the Taylor series expansion of $\sin(x)$.

Q: What happens to the tangential components of the tension?

The tangential components are $T\cos(d\theta)$ pointing in opposite directions along the tangent of the arc. Because the loop is symmetric and in equilibrium, these components perfectly cancel each other out, meaning there is no net force moving the wire along its own length.

Q: What would happen if the magnetic field was reversed?

If the magnetic field pointed out of the page, the magnetic force would act radially inwards. A flexible wire cannot sustain compressive forces (it has no structural rigidity), so instead of forming a circle, the loop would simply collapse inward.

Question

A thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

Options:

(A) $IBL$

(B) $\frac{IBL}{\pi}$

(C) $\frac{IBL}{2\pi}$

(D) $\frac{IBL}{4\pi}$

Sigma Percentile · Accepted Answer

This classic problem beautifully demonstrates the interplay between magnetic forces and mechanical tension in a flexible current-carrying wire. Let's break down the physics step-by-step to understand exactly why the wire forms a circle and how to calculate the tension within it.

### Analyzing the Setup

When a flexible wire carrying a current $I$ is placed in a uniform magnetic field $B$, every small segment of the wire experiences a magnetic force given by $d\vec{F} = I(d\vec{l} 	imes \vec{B})$.

Because the current is clockwise and the magnetic field points into the page, the right-hand rule tells us that the magnetic force on every segment acts **radially outwards**. This uniform outward "pressure" forces the flexible wire to expand until it takes the shape that maximizes its enclosed area for a given perimeter—a perfect circle.

Since the total length of the wire is $L$, the circumference of this newly formed circle must be $L$. We can easily find the radius $R$ of this circle:

$$2\pi R = L \implies R = \frac{L}{2\pi}$$

### The Master Equation

To find the tension $T$ in the wire, we need to analyze the forces acting on a very small differential element of the circular loop. Let's isolate a tiny segment that subtends a small angle $2d	heta$ at the center of the circle.

The length of this small segment is $dl = R(2d	heta)$. The outward magnetic force $F_m$ acting on this segment is:

$$F_m = I(dl)B = I(2R d	heta)B$$

This outward force must be perfectly balanced by the mechanical tension in the wire, otherwise, the wire would keep expanding. The tension $T$ acts tangentially at both ends of our small segment.

If we resolve these tension vectors into radial and tangential components, we find that the tangential components ($T \cos(d	heta)$) cancel each other out. However, the radial components point inwards towards the center. From each end of the segment, we get an inward force of $T \sin(d	heta)$.

Equating the total inward force to the outward magnetic force, we get:

$$2T \sin(d	heta) = F_m$$

### Final Calculation

Because we chose an infinitesimally small segment, the angle $d	heta$ is extremely small. In calculus, we use the small-angle approximation $\sin(d	heta) \approx d	heta$ (when measured in radians). Substituting this into our force balance equation gives:

$$2T(d	heta) = I(2R d	heta)B$$

Notice how elegantly the $2d	heta$ terms cancel out from both sides of the equation! This leaves us with a remarkably simple relationship for the tension:

$$T = IRB$$

Finally, we substitute the expression for the radius $R = \frac{L}{2\pi}$ that we found in the very first step:

$$T = I \left(\frac{L}{2\pi}ight) B = \frac{IBL}{2\pi}$$

This is our final answer. The tension in the wire is directly proportional to the current, the magnetic field, and the length of the wire.

Select Answer:

Visualized Solution

$L = 2 π R ⟹ R = \frac{L}{2 π}$

$Consider a small element subtending 2 d θ$

$F_{m} = I (d l) B = I (2 R d θ) B$

$2 T sin (d θ) = F_{m}$

$sin (d θ) \approx d θ ⟹ 2 T (d θ) = I (2 R d θ) B$

$T = I R B ⟹ T = \frac{I B L}{2 π}$

$What if B is reversed?$

The Sigma Insight: Magnetic Force on Current

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=2πR⟹R=2πL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider a small element subtending 2dθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fm​=I(dl)B=I(2Rdθ)B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2Tsin(dθ)=Fm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(dθ)≈dθ⟹2T(dθ)=I(2Rdθ)B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=IRB⟹T=2πIBL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if B is reversed?