A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a uniform

Q: Why doesn't the shape of the wire matter?

Because the magnetic force on a small element $d\mathbf{l}$ is $I(d\mathbf{l} \times \mathbf{B})$. When you integrate this over the entire wire in a *uniform* magnetic field, $\mathbf{B}$ is constant and can be pulled out of the integral. The integral of $d\mathbf{l}$ is simply the straight-line displacement vector $\mathbf{L}_{\text{eff}}$.

Q: What if the magnetic field was not uniform?

If the magnetic field varies with position, you cannot pull it out of the integral. In that case, the shape of the wire absolutely matters, and you would have to integrate $I(d\mathbf{l} \times \mathbf{B})$ along the exact path of the wire.

Q: How did we get the total length as $2(L+R)$?

By looking at the horizontal segments given in the diagram. The wire consists of a straight part $L$, an arc of width $R$, another straight part $R$, and a final arc of width $L$. Adding these horizontal projections gives $L + R + R + L = 2(L+R)$.

Q: Why is the force zero when $\mathbf{B}$ is along the x-axis?

The effective length vector is purely along the x-axis ($\hat{\mathbf{i}}$). If the magnetic field is also along the x-axis, the angle between them is zero. The cross product of two parallel vectors is zero, resulting in zero magnetic force.

Q: Does the current direction matter for the magnitude of the force?

No, reversing the current would only reverse the direction of the force vector (from $\hat{\mathbf{k}}$ to $-\hat{\mathbf{k}}$, for example), but the magnitude $F$ would remain exactly the same.

Question

A conductor (shown in the figure) carrying constant current $I$ is kept in the $x$-$y$ plane in a uniform magnetic field $\mathbf{B}$. If $F$ is the magnitude of the total magnetic force acting on the conductor, then the correct statements is/are

Options:

(A) if $\mathbf{B}$ is along $\hat{\mathbf{z}}$, $F \propto (L + R)$

(B) if $\mathbf{B}$ is along $\hat{\mathbf{x}}$, $F = 0$

(C) if $\mathbf{B}$ is along $\hat{\mathbf{y}}$, $F \propto (L + R)$

(D) if $\mathbf{B}$ is along $\hat{\mathbf{z}}$, $F = 0$

Sigma Percentile · Accepted Answer

### The Illusion of Shape

Imagine you are walking through a maze. You take twists, turns, and loops, but eventually, you reach the exit. If someone asks, "What is your net displacement?", you wouldn't describe every single turn. You would simply draw a straight line from the entrance to the exit.

In the beautiful world of electromagnetism, a uniform magnetic field behaves exactly like that person. It doesn't care about the complex twists and turns of a current-carrying wire. It only cares about the **effective length**—the straight-line displacement from the start to the end.

### The Master Equation

The magnetic force on a tiny, infinitesimal segment of wire $d\mathbf{l}$ is given by the Lorentz force law:
$$ d\mathbf{F} = I(d\mathbf{l} 	imes \mathbf{B}) $$

To find the total force on the entire wire, we integrate this expression:
$$ \mathbf{F} = \int I(d\mathbf{l} 	imes \mathbf{B}) $$

Here is where the magic happens. Because the magnetic field $\mathbf{B}$ is **uniform** (constant everywhere in space), we can pull it out of the integral!
$$ \mathbf{F} = I \left( \int d\mathbf{l} ight) 	imes \mathbf{B} $$

The integral of all the tiny displacement vectors $d\mathbf{l}$ is simply the total displacement vector from the starting point $P$ to the ending point $Q$. We call this the effective length vector, $\mathbf{L}_{	ext{eff}}$.
$$ \mathbf{F} = I(\mathbf{L}_{	ext{eff}} 	imes \mathbf{B}) $$

### Analyzing the Setup

Let's look at our specific wire. It starts at point $P$ and ends at point $Q$. Even though it has straight segments and curved arcs, we only need to find its total horizontal displacement.

Looking at the dimensions provided:
1. A straight segment of length $L$
2. An arc with a horizontal width of $R$
3. A straight segment of length $R$
4. An arc with a horizontal width of $L$

Adding these up, the total displacement along the x-axis is:
$$ \mathbf{L}_{	ext{eff}} = (L + R + R + L)\hat{\mathbf{i}} = 2(L+R)\hat{\mathbf{i}} $$

### Testing the Scenarios

Now, we simply evaluate the cross product for the different magnetic field directions given in the options.

**Case 1: Magnetic field along the x-axis**
If $\mathbf{B} = B\hat{\mathbf{i}}$, the force is:
$$ \mathbf{F} = I(2(L+R)\hat{\mathbf{i}} 	imes B\hat{\mathbf{i}}) $$
Since the cross product of any vector with itself is zero ($\hat{\mathbf{i}} 	imes \hat{\mathbf{i}} = 0$), the total force is **zero**. This makes option (b) correct.

**Case 2: Magnetic field along the y-axis**
If $\mathbf{B} = B\hat{\mathbf{j}}$, the force is:
$$ \mathbf{F} = I(2(L+R)\hat{\mathbf{i}} 	imes B\hat{\mathbf{j}}) = 2I(L+R)B(\hat{\mathbf{i}} 	imes \hat{\mathbf{j}}) $$
Since $\hat{\mathbf{i}} 	imes \hat{\mathbf{j}} = \hat{\mathbf{k}}$, the force is $\mathbf{F} = 2I(L+R)B\hat{\mathbf{k}}$. 
The magnitude is $F = 2I(L+R)B$, which means $F \propto (L+R)$. This makes option (c) correct.

**Case 3: Magnetic field along the z-axis**
If $\mathbf{B} = B\hat{\mathbf{k}}$, the force is:
$$ \mathbf{F} = I(2(L+R)\hat{\mathbf{i}} 	imes B\hat{\mathbf{k}}) = 2I(L+R)B(\hat{\mathbf{i}} 	imes \hat{\mathbf{k}}) $$
Since $\hat{\mathbf{i}} 	imes \hat{\mathbf{k}} = -\hat{\mathbf{j}}$, the force is $\mathbf{F} = -2I(L+R)B\hat{\mathbf{j}}$.
The magnitude is again $F = 2I(L+R)B$, which means $F \propto (L+R)$. This makes option (a) correct, and option (d) incorrect.

By trusting the math and focusing on the endpoints, a seemingly complex geometry problem collapses into a beautiful, straightforward vector calculation!

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Visualized Solution

$The Effective Length Concept$

$Calculating L_{eff}$

$Case 1: B along \hat{x}$

$Case 2: B along \hat{y}$

$Case 3: B along \hat{z}$

$Conclusion$

The Sigma Insight: Magnetic Force on Current

The Illusion of Shape

The Master Equation

Analyzing the Setup

Testing the Scenarios

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Effective Length Concept

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: B along x^

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: B along y^​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 3: B along z^

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