Two long parallel wires carrying currents 2.5 A and I (ampere) in the same direction (directed into

Q: Why do we add the magnetic fields of P and Q instead of subtracting them?

Because both wires carry current in the same direction (into the page) and point R is on the same side (right) of both wires, their magnetic fields at R point in the same direction (downwards).

Q: How do we know the direction of the force on the electron?

Using the cross product $\vec{v} \times \vec{B}$, velocity is $+x$ and field is $-y$, so the cross product is $-z$. Since the electron has a negative charge, the force is in the $+z$ direction (out of the page).

Q: Why is the angle $\theta$ in the force formula $90^\circ$?

The electron is moving along the x-axis, while the magnetic field is along the y-axis. The angle between the x and y axes is exactly $90^\circ$.

Q: For part (b), why are there two possible positions for the third wire?

A distance of 1 meter from R can be either 1 meter to the left or 1 meter to the right. Both positions can produce the required upward magnetic field if the current direction is chosen correctly.

Q: How do we determine the current direction for the third wire?

We need an upward magnetic field at R. Using the Right-Hand Grip Rule, if the wire is to the left of R, the thumb must point out of the page. If it's to the right, the thumb must point into the page.

Question

Two long parallel wires carrying currents 2.5 A and I (ampere) in the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of paper. The points P and Q are located at a distance of 5 m and 2 m respectively from a collinear point R (see figure).
(a) An electron moving with a velocity of $4 	imes 10^5$ m/s along the positive x-direction experiences a force of magnitude $3.2 	imes 10^{-20}$ N at the point R. Find the value of I.
(b) Find all the positions at which a third long parallel wire carrying a current of magnitude 2.5 A may be placed, so that the magnetic induction at R is zero.

Sigma Percentile · Accepted Answer

### Setting the Stage

Imagine a coordinate system where two infinitely long, parallel wires, $P$ and $Q$, are piercing through the plane of your paper. Wire $P$ sits right at the origin ($x = 0$), carrying a current of $2.5	ext{ A}$ into the page. Wire $Q$ is located at $x = 3	ext{ m}$, carrying an unknown current $I$, also into the page.

Our point of interest is $R$, located at $x = 5	ext{ m}$. Notice how the distances are perfectly laid out: $P$ is $5	ext{ m}$ away from $R$, and $Q$ is $2	ext{ m}$ away from $R$.

### Part A: The Invisible Force

First, we need to determine the magnetic environment at point $R$. Using the Right-Hand Grip Rule—pointing your right thumb into the page—your fingers will curl clockwise. At point $R$, which is to the right of both wires, the magnetic field vectors from both $P$ and $Q$ will point straight down (in the $-\hat{j}$ direction).

Let's calculate the net magnetic field $B_{	ext{net}}$ at $R$:

$$B_{	ext{net}} = B_P + B_Q = \frac{\mu_0 I_P}{2\pi r_P} + \frac{\mu_0 I_Q}{2\pi r_Q}$$

Substituting the known values:

$$B_{	ext{net}} = \frac{\mu_0}{2\pi} \left( \frac{2.5}{5} + \frac{I}{2} ight) = \frac{\mu_0}{4\pi} (1 + I)$$

Since $\frac{\mu_0}{4\pi} = 10^{-7}$, we get:

$$B_{	ext{net}} = 10^{-7} (1 + I) 	ext{ T}$$

Now, an electron zooms through point $R$ with a velocity $v = 4 	imes 10^5	ext{ m/s}$ along the positive x-axis. The magnetic force on a moving charge is given by the Lorentz force equation, $F_m = qvB \sin	heta$. Since the velocity is along the x-axis and the magnetic field is along the y-axis, the angle $	heta$ is exactly $90^\circ$.

$$F_m = e v B_{	ext{net}}$$

We are given that this force is $3.2 	imes 10^{-20}	ext{ N}$. Let's plug everything in:

$$3.2 	imes 10^{-20} = (1.6 	imes 10^{-19}) (4 	imes 10^5) 	imes 10^{-7} (1 + I)$$

$$3.2 	imes 10^{-20} = 6.4 	imes 10^{-21} (1 + I)$$

Dividing both sides by $6.4 	imes 10^{-21}$ gives us:

$$1 + I = 5 \implies \mathbf{I = 4	ext{ A}}$$

### Part B: Restoring Balance

Now that we know $I = 4	ext{ A}$, the net magnetic field at $R$ is:

$$B_{	ext{net}} = 10^{-7} (1 + 4) = 5 	imes 10^{-7}	ext{ T}$$

This field points downwards. To make the net magnetic field at $R$ exactly zero, we must introduce a third wire carrying $2.5	ext{ A}$ that produces an equal but opposite magnetic field—meaning it must produce $5 	imes 10^{-7}	ext{ T}$ pointing **upwards**.

Let's find the required distance $r$ for this third wire:

$$B_3 = \frac{\mu_0 I_3}{2\pi r} = 5 	imes 10^{-7}$$

$$\frac{4\pi 	imes 10^{-7} 	imes 2.5}{2\pi r} = 5 	imes 10^{-7}$$

$$\frac{5 	imes 10^{-7}}{r} = 5 	imes 10^{-7} \implies \mathbf{r = 1	ext{ m}}$$

### The Right-Hand Rule Magic

The third wire must be placed exactly $1	ext{ m}$ away from $R$. But wait, that gives us two possible locations on the x-axis: $1	ext{ m}$ to the left (at $x = 4	ext{ m}$) or $1	ext{ m}$ to the right (at $x = 6	ext{ m}$).

To ensure the magnetic field points upwards ($+\hat{j}$) at $R$:
- If placed at **$x = 4	ext{ m}$ (left of $R$)**, the Right-Hand Grip Rule tells us the current must flow **out of the page**.
- If placed at **$x = 6	ext{ m}$ (right of $R$)**, the current must flow **into the page**.

Both configurations perfectly cancel out the existing magnetic field, restoring absolute balance at point $R$!

Visualized Solution

\text{Analyzing the Setup}

\text{Magnetic Field at R due to P and Q}

\text{Net Magnetic Field at R}

\text{Magnetic Force on the Electron}

\text{Solving for Current } I

\text{Part (b): Nullifying the Magnetic Field}

\text{Distance of the Third Wire}

\text{Positions and Current Directions}

The Sigma Insight: Biot-Savart Law

Setting the Stage

Part A

Part B

The Right-Hand Rule Magic

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