A straight segment OC (of length L) of a circuit carrying a current I is placed along the x-axis. Tw

Q: Why can't we just use the formula F = ILB directly?

The formula $\mathbf{F} = I(\mathbf{L} \times \mathbf{B})$ is only valid when the magnetic field $\mathbf{B}$ is uniform over the entire length of the wire. Here, the magnetic field changes as you move along the $x$-axis, so you must integrate.

Q: How did we know the direction of the magnetic field from wire A?

Using the Right-Hand Grip Rule. Point your right thumb in the direction of the current (into the page). Your fingers curl clockwise. At a point on the $x$-axis, the tangent to this circular field points perpendicular to the line connecting the wire and the point.

Q: Why do the x-components of the magnetic field cancel out?

Because of symmetry. Wire $A$ is at $y = -a$ and wire $B$ is at $y = +a$. They carry the same current in the same direction. At any point on the $x$-axis, the field from $A$ points slightly right, and the field from $B$ points slightly left by the exact same amount.

Q: How did we integrate x / (x^2 + a^2)?

We used substitution. Let $t = x^2 + a^2$. Then $dt = 2x dx$, or $x dx = dt / 2$. The integral becomes $\frac{1}{2} \int \frac{dt}{t}$, which evaluates to $\frac{1}{2} \ln(t)$.

Q: Why is the force zero when the current in wire B is reversed?

When the current in $B$ is reversed, the symmetry changes. The $y$-components of the magnetic fields now cancel, and the net field is entirely along the $x$-axis. Since the wire segment $OC$ also lies on the $x$-axis, the current and the magnetic field are parallel. The cross product of parallel vectors is zero, so the force is zero.

Question

A straight segment $OC$ (of length $L$) of a circuit carrying a current $I$ is placed along the $x$-axis. Two infinitely long straight wires $A$ and $B$, each extending from $z = -\infty$ to $+\infty$, are fixed at $y = -a$ and $y = +a$ respectively, as shown in the figure. If the wires $A$ and $B$ each carry a current $I$ into the plane of the paper, obtain the expression for the force acting on the segment $OC$. What will be the force on $OC$ if the current in the wire $B$ is reversed?

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a straight wire segment $OC$ lying on the $x$-axis, carrying a current $I$. We also have two infinitely long wires, $A$ and $B$, positioned at $y = -a$ and $y = +a$, respectively. Both these wires carry a current $I$ into the plane of the screen. Our goal is to find the total magnetic force acting on the segment $OC$.

Since the magnetic field varies along the $x$-axis, we can't just use a simple formula like $\mathbf{F} = I(\mathbf{L} 	imes \mathbf{B})$. Therefore, we'll consider a tiny element of length $dx$ on the segment $OC$, located at a distance $x$ from the origin.

### The Magnetic Field at an Arbitrary Point
First, let's determine the magnetic field at this point $P$ due to the two infinite wires. Using the right-hand grip rule, the magnetic field $\mathbf{B}_A$ due to wire $A$ points perpendicular to the line $AP$. Similarly, the magnetic field $\mathbf{B}_B$ due to wire $B$ points perpendicular to the line $BP$.

Since both wires are at the same distance from $P$ and carry the same current, the magnitudes of $\mathbf{B}_A$ and $\mathbf{B}_B$ are equal:
$$|\mathbf{B}_A| = |\mathbf{B}_B| = \frac{\mu_0 I}{2\pi \sqrt{x^2 + a^2}}$$

If we resolve these magnetic field vectors into components, notice that their $x$-components are equal and opposite, so they perfectly cancel each other out! However, their $y$-components point in the same direction, along the negative $y$-axis. They add up to give the net magnetic field, $\mathbf{B}_{net}$, which equals $2B \cos	heta$.

$$B_{net} = 2 \left( \frac{\mu_0 I}{2\pi \sqrt{x^2 + a^2}} ight) \left( \frac{x}{\sqrt{x^2 + a^2}} ight) = \frac{\mu_0 I x}{\pi (x^2 + a^2)}$$

So, $\mathbf{B}_{net} = -\frac{\mu_0 I x}{\pi (x^2 + a^2)} \hat{\mathbf{j}}$.

### Calculating the Force
So, what is the force on our tiny element $dx$? The magnetic force $d\mathbf{F}$ is given by $I(d\mathbf{x} 	imes \mathbf{B})$. Since the current element is along the positive $x$-axis and the magnetic field is along the negative $y$-axis, the cross product gives a force in the negative $z$-direction, into the screen.

$$d\mathbf{F} = I(dx \hat{\mathbf{i}}) 	imes \left( -\frac{\mu_0 I x}{\pi (x^2 + a^2)} \hat{\mathbf{j}} ight) = -\frac{\mu_0 I^2 x}{\pi (x^2 + a^2)} dx \hat{\mathbf{k}}$$

To find the total force on the entire segment $OC$, we integrate this expression from $x = 0$ to $x = L$. We can solve this integral easily by substituting $x^2 + a^2 = t$, which gives $2x dx = dt$. The integration yields a natural logarithm.

$$\mathbf{F} = \int_0^L -\frac{\mu_0 I^2 x}{\pi (x^2 + a^2)} dx \hat{\mathbf{k}} = -\frac{\mu_0 I^2}{2\pi} \left[ \ln(x^2 + a^2) ight]_0^L \hat{\mathbf{k}}$$

Plugging in the limits, we get the final expression for the force:

**$$\mathbf{F} = -\frac{\mu_0 I^2}{2\pi} \ln\left(\frac{L^2 + a^2}{a^2}ight) \hat{\mathbf{k}}$$**

### Reversing the Current in Wire B
Now let's look at the second part of the question. What happens if the current in wire $B$ is reversed?

If wire $B$ carries current out of the page, the direction of $\mathbf{B}_B$ flips. Now, the $y$-components cancel out, and the net magnetic field points entirely along the $x$-axis ($-\hat{\mathbf{i}}$).

Since the net magnetic field is now parallel to the current segment $OC$, the cross product of $d\mathbf{x}$ and $\mathbf{B}$ becomes zero. Therefore, the magnetic force acting on the segment $OC$ will be exactly **zero**! A beautiful result of symmetry.

Visualized Solution

Visualizing the Setup

Choosing an Elemental Segment

Magnetic Field due to Wires $A$ and $B$

Net Magnetic Field at $P$

Force on the Elemental Segment

Integrating for Total Force

Final Force Expression

Reversing Current in Wire $B$

Force with Reversed Current

The Sigma Insight: Magnetic Force on Current

Analyzing the Setup Imagine a straight wire segment $O C$ lying on the $x$ -axis, carrying a current $I$

The Magnetic Field at an Arbitrary Point First, let's determine the magnetic field at this point $P$ due to the two infinite wires

Calculating the Force So, what is the force on our tiny element $d x$ ? The magnetic force $d F$ is given by $I (d x \times B)$

Reversing the Current in Wire B Now let's look at the second part of the question

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