A circular loop of radius R is bent along a diameter and given a shape as shown in figure. One of th

Q: Why isn't the net force on the loop zero in part (b)?

The net force on a *closed* loop in a uniform magnetic field is zero. However, in this problem, the current enters at K and leaves at M, splitting into two parallel branches. It is not a single closed loop, so the net force is non-zero.

Q: How do we determine the direction of the magnetic field for each semicircle?

Use the right-hand grip rule. Curl the fingers of your right hand in the direction of the current along the semicircle. Your extended thumb will point in the direction of the magnetic field at the center.

Q: Why is the displacement vector $\mathbf{l}$ the same for both semicircles?

The displacement vector is defined as the straight-line vector from the starting point to the ending point of the current path. Since both semicircles start at K and end at M, they share the exact same displacement vector.

Q: What happens to the $\hat{\mathbf{i}} \times \hat{\mathbf{i}}$ term in the cross product?

The cross product of any vector with itself (or a parallel vector) is always zero. This physically means that a magnetic field component parallel to the velocity does not exert any force on the particle.

Q: Does the shape of the wire matter at all in a uniform magnetic field?

No! As long as the magnetic field is perfectly uniform, the specific path the wire takes between two points doesn't affect the total magnetic force. Only the straight-line distance and direction between the endpoints matter.

Question

A circular loop of radius $R$ is bent along a diameter and given a shape as shown in figure. One of the semicircles ($KNM$) lies in the $x$-$z$ plane and the other one ($KLM$) in the $y$-$z$ plane with their centres at origin. Current $I$ is flowing through each of the semicircles as shown in figure.
(a) A particle of charge $q$ is released at the origin with a velocity $\mathbf{v} = -v_0 \hat{\mathbf{i}}$. Find the instantaneous force $\mathbf{F}$ on the particle. Assume that space is gravity free.
(b) If an external uniform magnetic field $B_0 \hat{\mathbf{j}}$ is applied determine the force $\mathbf{F}_1$ and $\mathbf{F}_2$ on the semicircles $KLM$ and $KNM$ due to the field and the net force $\mathbf{F}$ on the loop.

Sigma Percentile · Accepted Answer

### Visualizing the 3D Circuit

Look closely at this setup. We are dealing with a fascinating 3D geometry where a circular loop is bent along its diameter. This creates two distinct semicircular branches: one lying in the $y$-$z$ plane (let's call it $KLM$) and the other in the $x$-$z$ plane (let's call it $KNM$).

A critical detail to notice from the arrows in the diagram is the direction of the current. The current $I$ does not flow in a single continuous circle. Instead, it enters at point $K$ (located at $z = R$) and splits, flowing through both semicircular branches to exit at point $M$ (located at $z = -R$).

### The Superposition of Magnetic Fields

To find the instantaneous force on a moving charge at the origin, we first need to determine the net magnetic field $\mathbf{B}$ at that exact point. We can do this by applying the Biot-Savart law (or the standard formula for a semicircle) to each branch independently and using the principle of superposition.

First, let's analyze the $KLM$ loop in the $y$-$z$ plane. The current flows from $K$ through $L$ to $M$. If you curl the fingers of your right hand along this path, your thumb points directly along the negative $x$-axis. Therefore, the magnetic field contribution from this branch is:

$$\mathbf{B}_{KLM} = \frac{\mu_0 I}{4R} (-\hat{\mathbf{i}})$$

Next, look at the $KNM$ loop in the $x$-$z$ plane. Here, the current flows from $K$ through $N$ to $M$. Applying the right-hand rule again, your thumb points along the positive $y$-axis. Thus, its contribution is:

$$\mathbf{B}_{KNM} = \frac{\mu_0 I}{4R} (\hat{\mathbf{j}})$$

The net magnetic field at the origin is simply the vector sum of these two fields:

$$\mathbf{B} = \frac{\mu_0 I}{4R} (-\hat{\mathbf{i}} + \hat{\mathbf{j}})$$

### The Lorentz Force in Action

Now for part (a). A particle with charge $q$ is released at the origin with a velocity $\mathbf{v} = -v_0 \hat{\mathbf{i}}$. To find the instantaneous force, we use the Lorentz force equation:

$$\mathbf{F} = q(\mathbf{v} 	imes \mathbf{B})$$

Substitute our known vectors into the equation:

$$\mathbf{F} = q(-v_0 \hat{\mathbf{i}}) 	imes \frac{\mu_0 I}{4R} (-\hat{\mathbf{i}} + \hat{\mathbf{j}})$$

When we distribute the cross product, remember that the cross product of any vector with itself is zero ($\hat{\mathbf{i}} 	imes \hat{\mathbf{i}} = 0$). We only need to evaluate the cross term:

$$\mathbf{F} = -\frac{\mu_0 q v_0 I}{4R} (\hat{\mathbf{i}} 	imes \hat{\mathbf{j}})$$

Since $\hat{\mathbf{i}} 	imes \hat{\mathbf{j}} = \hat{\mathbf{k}}$, the final force on the particle is:

$$\mathbf{F} = -\frac{\mu_0 q v_0 I}{4R} \hat{\mathbf{k}}$$

### The Magic of the Displacement Vector

Moving to part (b), we are asked to find the force on the loop when an external uniform magnetic field $\mathbf{B}_{ext} = B_0 \hat{\mathbf{j}}$ is applied.

There is a beautiful shortcut in physics for this exact scenario. Whenever a curved, current-carrying wire is placed in a **uniform** magnetic field, the total magnetic force on it depends only on the straight displacement vector $\mathbf{l}$ connecting its starting and ending points. The exact path the wire takes does not matter!

Both semicircular loops start at $K(0, 0, R)$ and end at $M(0, 0, -R)$. Therefore, they share the exact same effective displacement vector:

$$\mathbf{l} = \mathbf{r}_M - \mathbf{r}_K = -2R \hat{\mathbf{k}}$$

We can now calculate the force on each individual branch using $\mathbf{F} = I(\mathbf{l} 	imes \mathbf{B}_{ext})$:

$$\mathbf{F}_1 = \mathbf{F}_2 = I(-2R \hat{\mathbf{k}} 	imes B_0 \hat{\mathbf{j}})$$

Since $\hat{\mathbf{k}} 	imes \hat{\mathbf{j}} = -\hat{\mathbf{i}}$, the negative signs cancel out, giving us:

$$\mathbf{F}_1 = \mathbf{F}_2 = 2B_0 I R \hat{\mathbf{i}}$$

Finally, the total net force on the entire structure is the sum of the forces on the two branches:

$$\mathbf{F}_{net} = \mathbf{F}_1 + \mathbf{F}_2 = 4B_0 I R \hat{\mathbf{i}}$$

Visualized Solution

$Visualizing the 3D Circuit$

$Magnetic Field of K L M$

$Magnetic Field of K N M$

$Net Magnetic Field$

$Particle Velocity$

$Lorentz Force Calculation$

$Effective Displacement Vector$

$Force due to External Field$

The Sigma Insight: Magnetic Force on Current

Visualizing the 3D Circuit

The Superposition of Magnetic Fields

The Lorentz Force in Action

The Magic of the Displacement Vector

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the 3D Circuit

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force due to External Field