As shown in the figure, two infinitely long, identical wires are bent by 90 degrees and placed in su

Q: Why is the magnetic field from segments LP and QM exactly zero?

Because the point O lies directly on the extended line of these segments. In the Biot-Savart Law, the magnetic field depends on the cross product $d\vec{l} \times \vec{r}$. If the point is on the axis, the angle between the current element and the position vector is $0^\circ$ or $180^\circ$, making the cross product (and thus the magnetic field) zero.

Q: What exactly is a semi-infinite wire?

A semi-infinite wire is a theoretical construct that starts at a specific point and extends to infinity in only one direction. The magnetic field at a point directly adjacent to its starting end is exactly half of the magnetic field produced by a fully infinitely long wire.

Q: How do I apply the Right-Hand Thumb Rule for segment PS?

Point your right thumb in the direction of the current in segment PS, which is upwards. Now, curl your fingers towards the origin O. You will notice your fingers are pointing directly into the page when they pass through O.

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

Magnetic field is a vector quantity. When you apply the Right-Hand Rule to both segment PS and segment QN, you will find that both of their magnetic fields at the origin point in the exact same direction (into the page). Therefore, their magnitudes add up algebraically.

Q: What would happen if the current in segment QN was flowing upwards instead?

If the current in QN was flowing upwards, applying the Right-Hand Rule would show its magnetic field at O pointing OUT of the page. Since the field from PS points INTO the page, the two fields would perfectly cancel each other out, resulting in a net magnetic field of zero at the origin!

Question

As shown in the figure, two infinitely long, identical wires are bent by $90^{\circ}$ and placed in such a way that the segments $LP$ and $QM$ are along the X-axis, while segments $PS$ and $QN$ are parallel to the Y-axis. If $OP = OQ = 4	ext{ cm}$ and the magnitude of the magnetic field at $O$ is $10^{-4}	ext{ T}$ and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at $O$ will be (Take, $\mu_0 = 4\pi 	imes 10^{-7}	ext{ NA}^{-2}$)

Options:

(A) 40 A, perpendicular out of the page

(B) 20 A, perpendicular into the page

(C) 20 A, perpendicular out of the page

(D) 40 A, perpendicular into the page

Sigma Percentile · Accepted Answer

Imagine you are standing at the origin $O$, surrounded by two infinitely long wires that have been bent at perfect right angles. This problem might look like a complex web of magnetic fields, but it is actually a beautiful exercise in symmetry and the Biot-Savart Law. Let's break it down piece by piece.

### Decoding the Geometry

We have two wires. Wire 1 comes from negative infinity along the x-axis, reaches point $P$, and then shoots straight up parallel to the y-axis towards positive infinity. Wire 2 comes from point $Q$ on the positive x-axis, goes to positive infinity along the x-axis, and also shoots straight down from $Q$ parallel to the negative y-axis.

Our goal is to find the magnetic field exactly at the origin $O$. To do this, we must treat each straight segment of the wires independently and then use the principle of superposition to find the net field.

### The Magic of the Axis

Let's first look at the horizontal segments: $LP$ and $QM$. Notice something special? The origin $O$ lies exactly on the extended line of both these segments.

According to the Biot-Savart Law, the magnetic field $d\vec{B}$ produced by a current element $d\vec{l}$ is proportional to $d\vec{l} 	imes \vec{r}$. If the point of interest lies on the axis of the current element, the angle between $d\vec{l}$ and $\vec{r}$ is either $0^\circ$ or $180^\circ$. In both cases, the cross product is zero! Therefore, **the magnetic field produced by segments $LP$ and $QM$ at the origin $O$ is exactly zero.**

### The Semi-Infinite Contributors

This massive simplification means that the entire magnetic field at $O$ is generated solely by the vertical segments: $PS$ and $QN$.

Because these segments start at the x-axis and extend to infinity, they act as **semi-infinite wires**. The magnetic field at a perpendicular distance $r$ from one end of a semi-infinite wire is exactly half the field of an infinitely long wire:

$$B = \frac{\mu_0 I}{4\pi r}$$

### The Right-Hand Rule in Action

Now, we must determine the direction of the magnetic field from each segment using the **Right-Hand Thumb Rule**.

For segment $PS$, the current flows upwards ($+\hat{j}$). If you point your right thumb up and curl your fingers towards the origin $O$ (which is to the right of the wire), your fingers will curl **into the page**.

For segment $QN$, the current flows downwards ($-\hat{j}$). Point your right thumb down and curl your fingers towards the origin $O$ (which is to the left of the wire). Once again, your fingers will curl **into the page**.

### Bringing It All Together

Since both $\vec{B}_{PS}$ and $\vec{B}_{QN}$ point in the exact same direction (perpendicularly into the page), their magnitudes simply add up algebraically:

$$B_{net} = B_{PS} + B_{QN} = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{2\pi r}$$

We are given that the net magnetic field $B_{net} = 10^{-4}	ext{ T}$ and the distance $r = OP = OQ = 4	ext{ cm} = 0.04	ext{ m}$. Substituting these values into our master equation:

$$10^{-4} = \frac{4\pi 	imes 10^{-7} 	imes I}{2\pi 	imes 0.04}$$

$$10^{-4} = \frac{2 	imes 10^{-7} 	imes I}{0.04}$$

Rearranging to solve for the current $I$:

$$I = \frac{10^{-4} 	imes 0.04}{2 	imes 10^{-7}} = \frac{4 	imes 10^{-6}}{2 	imes 10^{-7}} = 20	ext{ A}$$

Thus, the current in each wire is **$20	ext{ A}$**, and the net magnetic field is directed **perpendicularly into the page**.

Select Answer:

Visualized Solution

System Setup

Field due to Axial Segments

Field due to Semi-Infinite Segments

Direction of $B_{P S}$

Direction of $B_{QN}$

Net Magnetic Field

Substituting Values

Final Calculation

The Sigma Insight: Biot-Savart Law

Decoding the Geometry

The Magic of the Axis

The Semi-Infinite Contributors

The Right-Hand Rule in Action

Bringing It All Together

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