A current of 10 A flows around a closed path in a circuit which is in the horizontal plane as shown

Q: Why do the radial segments not produce any magnetic field at the center?

According to the Biot-Savart Law, $d\mathbf{B} \propto d\mathbf{l} \times \mathbf{r}$. For radial segments, the current element $d\mathbf{l}$ is parallel (or anti-parallel) to the position vector $\mathbf{r}$ pointing to the center. The cross product of parallel vectors is zero.

Q: How did we know the total angle of the inner arcs is $180^\circ$?

The circuit is a closed loop with 8 alternating arcs (4 inner, 4 outer) that subtend equal angles. Since a full circle is $360^\circ$, each arc subtends $360^\circ / 8 = 45^\circ$. Four inner arcs combined subtend $4 \times 45^\circ = 180^\circ$.

Q: Why is the force on the central wire zero?

The magnetic field produced by the horizontal circuit at its center points vertically (along the z-axis). The central wire is also placed vertically. Since the current in the wire is parallel to the magnetic field, the cross product in the magnetic force formula is zero.

Q: Why is the force on arc AC zero?

The magnetic field produced by a long straight central wire forms concentric circles. At the position of arc AC, this magnetic field is tangential to the arc. The current in arc AC also flows tangentially. Since they are parallel, the magnetic force is zero.

Q: Why did we have to integrate to find the force on segment CD?

Segment CD is a radial line extending from $r_1$ to $r_2$. The magnetic field from the central wire is not uniform; it decreases with distance as $B = \frac{\mu_0 I}{2\pi x}$. Because the field varies along the length of the segment, we must integrate the force over infinitesimally small elements $dx$.

Question

A current of 10 A flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii $r_1 = 0.08$ m and $r_2 = 0.12$ m. Each subtends the same angle at the centre.
(a) Find the magnetic field produced by this circuit at the centre.
(b) An infinitely long straight wire carrying a current of 10 A is passing through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. What is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the arc $AC$ and the straight segment $CD$ due to the current at the centre?

Sigma Percentile · Accepted Answer

This problem is a beautiful exercise in applying the Biot-Savart Law and understanding the vector nature of magnetic forces. It tests your ability to break down a complex geometry into simpler, manageable parts.

### Analyzing the Geometry of the Circuit

Look closely at the circuit. It consists of eight alternating arcs—four on the inside with radius $r_1$ and four on the outside with radius $r_2$. These arcs are connected by straight radial lines.

Since the circuit is divided into eight equal sectors, each arc subtends an angle of $\frac{360^\circ}{8} = 45^\circ$ at the center. If we sum the angles of the four inner arcs, we get $4 	imes 45^\circ = 180^\circ$, which is exactly $\pi$ radians. This means the four inner arcs collectively contribute to the magnetic field exactly as a single semicircle of radius $r_1$ would. The same logic applies to the four outer arcs, which act like a semicircle of radius $r_2$.

### The Magnetic Field at the Center

First, let's consider the straight radial segments. According to the Biot-Savart Law, the magnetic field contribution is proportional to $d\mathbf{l} 	imes \mathbf{r}$. For any radial segment, the current element $d\mathbf{l}$ is parallel (or anti-parallel) to the position vector $\mathbf{r}$ pointing towards the center. The cross product of parallel vectors is zero, meaning **the straight radial wires contribute absolutely nothing to the magnetic field at the center**.

The total magnetic field is therefore entirely due to the arcs. Using the formula for the magnetic field at the center of a circular arc, $B = \frac{\mu_0 i}{4\pi r} 	heta$, we can write:

$$B_{	ext{inner}} = \frac{\mu_0 i}{4\pi r_1} (\pi) = \frac{\mu_0 i}{4 r_1}$$

$$B_{	ext{outer}} = \frac{\mu_0 i}{4\pi r_2} (\pi) = \frac{\mu_0 i}{4 r_2}$$

By the Right-Hand Rule, since the current flows anti-clockwise, both of these fields point outwards (out of the plane of the paper). We can simply add them up:

$$B_{	ext{net}} = B_{	ext{inner}} + B_{	ext{outer}} = \frac{\mu_0 i}{4} \left( \frac{1}{r_1} + \frac{1}{r_2} ight)$$

Substituting the given values ($i = 10 	ext{ A}$, $r_1 = 0.08 	ext{ m}$, $r_2 = 0.12 	ext{ m}$):

$$B_{	ext{net}} = \frac{4\pi 	imes 10^{-7} 	imes 10}{4\pi} 	imes \pi \left( \frac{0.08 + 0.12}{0.08 	imes 0.12} ight)$$

$$B_{	ext{net}} = 6.54 	imes 10^{-5} 	ext{ T}$$

### The Interaction with the Central Wire

Now, imagine we place an infinitely long straight wire right at the center, carrying $I = 10 	ext{ A}$ into the paper. What is the force on this wire?

The magnetic force on a straight wire is given by $\mathbf{F} = I (\mathbf{L} 	imes \mathbf{B})$. The magnetic field produced by our circuit points vertically outwards, while the current in the central wire flows vertically inwards. Because the length vector $\mathbf{L}$ and the magnetic field $\mathbf{B}_{	ext{net}}$ are anti-parallel, their cross product is zero. **Therefore, the force on the central wire is zero.**

### Force on the Arc AC

Next, we need to find the force exerted *by* the central wire *on* the arc AC. The magnetic field of a long straight wire forms concentric circles. At the location of arc AC, this magnetic field is perfectly tangential to the arc.

The current in arc AC also flows tangentially along the arc. Once again, the current element $d\mathbf{l}$ and the magnetic field $\mathbf{B}_{	ext{wire}}$ are parallel. The cross product $d\mathbf{l} 	imes \mathbf{B}_{	ext{wire}}$ vanishes, meaning **absolutely no magnetic force acts on arc AC**.

### Integrating the Force on Segment CD

Finally, let's look at segment CD. This is a radial line extending from $r_1$ to $r_2$. The magnetic field from the central wire is perpendicular to this radial segment!

However, the magnetic field is not uniform; it decreases with distance $x$ from the center as $B_{	ext{wire}} = \frac{\mu_0 I}{2\pi x}$. To find the total force, we must integrate the force over infinitesimally small elements $dx$:

$$dF = i (dx) B_{	ext{wire}} = i \left( \frac{\mu_0 I}{2\pi x} ight) dx$$

Integrating from $r_1$ to $r_2$:

$$F_{CD} = \int_{r_1}^{r_2} \frac{\mu_0 i I}{2\pi x} dx = \frac{\mu_0 i I}{2\pi} \ln\left(\frac{r_2}{r_1}ight)$$

Substituting the numbers:

$$F_{CD} = \frac{4\pi 	imes 10^{-7} 	imes 10 	imes 10}{2\pi} \ln\left(\frac{0.12}{0.08}ight)$$

$$F_{CD} = 2 	imes 10^{-5} 	imes \ln(1.5)$$

**$$F_{CD} = 8.1 	imes 10^{-6} 	ext{ N}$$**

By Fleming's Left-Hand Rule, this force points inwards, perpendicular to the segment CD.

Visualized Solution

$B_{center}$ Setup

Field from Radial Segments

Field from Arcs

Total Magnetic Field

Central Wire Introduced

Force on Arc AC

Force on Segment CD

Final Calculation for CD

The Sigma Insight: Biot-Savart Law

Analyzing the Geometry of the Circuit

The Magnetic Field at the Center

The Interaction with the Central Wire

Force on the Arc AC

Integrating the Force on Segment CD

Similar Questions

A very long wire ABDMNDC is shown in figure carrying current $I$ . $A B$ and $B C$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $D M N D$ of radius $R$ . $A B$ , $B C$ parts are tangential to circular turn at $N$ and $D$ . Magnetic field at the centre of circle is

A current loop, having two circular arcs joined by two radial lines as shown in the figure. It carries a current of $10 A$ . The magnetic field at point $O$ will be close to

Find the magnetic field at point $P$ due to a straight line segment $A B$ of length $6 cm$ carrying a current of $5 A$ (See figure). (Take, $μ_{0} = 4 π \times 1 0^{- 7} N-A^{- 2}$ )

A non-planar loop of conducting wire carrying a current $I$ is placed as shown in the figure. Each of the straight sections of the loop is of length $2 a$ . The magnetic field due to this loop at the point $P (a, 0, a)$ points in the direction

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is [Take, $μ_{0} = 4 π \times 1 0^{- 7} NA^{- 2}$ ]

A current of $1.5 A$ is flowing through a triangle, of side $9 cm$ each. The magnetic field at the centroid of the triangle is (Assume that, the current is flowing in the clockwise direction.)

A long straight wire along the z-axis carries a current $I$ in the negative z-direction. The magnetic vector field $B$ at a point having coordinate $(x, y)$ on the $z = 0$ plane is