Two very long straight parallel wires carry steady currents I and -I respectively. The distance betw

Q: Why do the magnetic fields from the two wires add up instead of canceling out?

Because the currents are in opposite directions and the point is between them. Using the Right-Hand Grip Rule, the field from the first wire points 'up' (or 'down') at the midpoint, and the field from the second wire also points in the exact same direction. They reinforce each other!

Q: What if the velocity was in the plane of the wires?

If the velocity was in the plane of the wires, it would be perpendicular to the magnetic field (which is perpendicular to the plane). In that case, the angle $\theta$ would be $90^\circ$, and the force would be maximum: $F = qvB_{net}$.

Q: Does the distance $d$ matter for the final answer?

Not at all! While $d$ determines the magnitude of the magnetic field $B_{net}$, the force is zero purely because of the cross product $\mathbf{v} \times \mathbf{B} = 0$. The geometry of the vectors makes the magnitude of $B_{net}$ irrelevant.

Q: Could the velocity be parallel instead of anti-parallel?

Yes! The problem only states that the velocity is perpendicular to the plane. It could point 'up' (parallel to $\mathbf{B}$) or 'down' (anti-parallel to $\mathbf{B}$). In both cases, $\sin(0^\circ) = 0$ and $\sin(180^\circ) = 0$, so the force is zero regardless.

Q: Is this a trick question?

In a way, yes! It's a classic JEE conceptual trap. They give you all the variables ($I$, $d$, $q$, $v$) to tempt you into calculating the exact magnitude of the magnetic field and plugging it into a formula. But if you check the vector directions first, you save all that time!

Question

Two very long straight parallel wires carry steady currents $I$ and $-I$ respectively. The distance between the wires is $d$. At a certain instant of time, a point charge $q$ is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity $\mathbf{v}$ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

Options:

(A) $\frac{\mu_0 Iqv}{2\pi d}$

(B) $\frac{\mu_0 Iqv}{\pi d}$

(C) $\frac{2\mu_0 Iqv}{\pi d}$

(D) zero

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine you are looking at a flat surface, let's call it the $xy$-plane. On this plane, there are two infinitely long, straight wires running parallel to each other. They are separated by a distance $d$.

The first wire carries a steady current $I$, and the second wire carries a current $-I$. This means the currents are flowing in exactly opposite directions. Right in the middle of these two wires, at a distance of $d/2$ from each, sits a point charge $q$.

### The Magnetic Field from the Wires

To find the force on the charge, we first need to know the magnetic field at its location. Let's use the Right-Hand Grip Rule.

Point your right thumb in the direction of the current $I$ in the first wire. Your fingers will curl around the wire. At the midpoint between the wires, your fingers will be pointing straight up, perpendicular to the plane of the wires. So, the magnetic field $\mathbf{B}_1$ from the first wire points upwards.

Now, let's look at the second wire. It carries a current $-I$ (opposite direction) and is located on the opposite side of the midpoint. If you apply the Right-Hand Grip Rule again, you'll find that its magnetic field $\mathbf{B}_2$ at the midpoint *also* points straight up!

Since both $\mathbf{B}_1$ and $\mathbf{B}_2$ point in the exact same direction, they reinforce each other. The net magnetic field $\mathbf{B}_{net}$ is simply their sum, and it points strictly perpendicular to the plane of the wires.

### The Lorentz Force and the Catch

Now, let's bring the charge $q$ into motion. The problem states that its instantaneous velocity $\mathbf{v}$ is perpendicular to the plane of the wires.

Wait a minute... the net magnetic field $\mathbf{B}_{net}$ is perpendicular to the plane, and the velocity $\mathbf{v}$ is *also* perpendicular to the plane! This means that the velocity vector and the magnetic field vector lie along the exact same line. They must be either parallel ($	heta = 0^\circ$) or anti-parallel ($	heta = 180^\circ$).

The magnetic force on a moving charge is given by the Lorentz force equation:

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

The magnitude of this force is:

$$|\mathbf{F}| = qvB_{net} \sin	heta$$

### The Final Verdict

Here is the beautiful catch of this problem. Because the velocity and the magnetic field are collinear, the angle $	heta$ between them is either $0^\circ$ or $180^\circ$.

In trigonometry, we know that $\sin(0^\circ) = 0$ and $\sin(180^\circ) = 0$.

Therefore, regardless of the exact magnitude of the magnetic field or the speed of the charge, the cross product vanishes completely.

$$|\mathbf{F}| = qvB_{net} (0) = 0$$

The magnitude of the magnetic force acting on the charge at this instant is exactly **zero**.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$B_{1} ⊥ Plane$

$B_{2} ⊥ Plane$

$B_{n e t} = B_{1} + B_{2}$

$v ⊥ Plane$

$θ = 18 0^{\circ}$

$∣ F ∣ = q v B_{n e t} sin θ$

$∣ F ∣ = 0$

$What if v ∥ Plane?$

The Sigma Insight: Motion of a Charge in Magnetic Fields

Visualizing the Setup

The Magnetic Field from the Wires

The Lorentz Force and the Catch

The Final Verdict

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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; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B1​⊥Plane

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2​⊥Plane

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​=B1​+B2​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=180∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣F∣=qvBnet​sinθ

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

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