Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB ca

Q: Why is the magnetic field of wire AOB along the y-axis?

By the right-hand thumb rule, if you point your thumb along the current (x-axis), your fingers curl around it. At a point on the positive z-axis, the tangent to this curl points in the positive y-direction.

Q: Can we just add the magnitudes $B_1$ and $B_2$?

No, magnetic field is a vector quantity. Since they point in different directions (y and -x), you must add them vectorially.

Q: What if the wires were not infinitely long?

If the wires were finite, you would use the formula for a finite wire: $B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$. The directions would remain the same, but the magnitudes would be smaller.

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

Reversing a current would reverse its magnetic field vector. However, since the two vectors will always be perpendicular regardless of their signs, the magnitude of the resultant $\sqrt{B_1^2 + B_2^2}$ remains unchanged.

Q: Why did we use the formula for an infinite wire?

The problem states 'Two identical conducting wires...'. In JEE context, unless specified otherwise, long straight wires are treated as infinitely long to avoid edge effects.

Question

Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current $I_1$ and COD carries a current $I_2$. The magnetic field on a point lying at a distance $d$ from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by

Options:

(A) $\frac{\mu_0}{2\pi}\left(\frac{I_1+I_2}{d}\right)^{1/2}$

(B) $\frac{\mu_0}{2\pi d}(I_1^2+I_2^2)^{1/2}$

(C) $\frac{\mu_0}{2\pi d}(I_1+I_2)$

(D) $\frac{\mu_0}{2\pi d}(I_1^2+I_2^2)$

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine two infinitely long, straight conducting wires crossing each other at a perfect right angle. Let's place the first wire, AOB, along the x-axis, carrying a steady current $I_1$. The second wire, COD, lies along the y-axis, carrying a current $I_2$. They intersect at the origin, O.

Our goal is to find the net magnetic field at a specific point P. This point is located at a distance $d$ from the origin, directly along the z-axis. Because the z-axis is perpendicular to both the x and y axes, point P lies in a plane perpendicular to the plane containing the two wires.

### The Magnetic Field of a Single Wire

To solve this, we first need to recall the fundamental formula for the magnetic field produced by an infinitely long, straight current-carrying wire. According to Ampere's Law (or derived from the Biot-Savart Law), the magnitude of the magnetic field $B$ at a perpendicular distance $r$ from a wire carrying current $I$ is given by:

$$B = \frac{\mu_0 I}{2\pi r}$$

We will apply this principle independently to both wires to find their individual contributions to the magnetic field at point P.

### Analyzing the Individual Fields

Let's start with wire AOB. It carries current $I_1$ along the x-axis. Point P is at a distance $d$ along the z-axis. Using the formula, the magnitude of the magnetic field $\vec{B}_1$ at point P is:

$$B_1 = \frac{\mu_0 I_1}{2\pi d}$$

But what about its direction? Using the right-hand thumb rule—pointing your thumb in the direction of the current (positive x-axis) and curling your fingers towards point P (positive z-axis)—your fingers will point in the direction of the positive y-axis. So, $\vec{B}_1$ is directed along the +y-axis.

Now, let's look at wire COD. It carries current $I_2$ along the y-axis. The magnitude of its magnetic field $\vec{B}_2$ at point P is:

$$B_2 = \frac{\mu_0 I_2}{2\pi d}$$

Applying the right-hand thumb rule again—thumb along the positive y-axis, curling towards the positive z-axis—your fingers will point in the direction of the negative x-axis. Thus, $\vec{B}_2$ is directed along the -x-axis.

### Vector Addition for the Net Field

We now have two magnetic field vectors at point P: $\vec{B}_1$ pointing along the y-axis, and $\vec{B}_2$ pointing along the negative x-axis. Because the x and y axes are perpendicular, these two magnetic field vectors are also perfectly perpendicular to each other.

Since magnetic field is a vector quantity, the net magnetic field $\vec{B}_{net}$ is the vector sum of $\vec{B}_1$ and $\vec{B}_2$. For two perpendicular vectors, the magnitude of their resultant is given by the Pythagorean theorem:

$$B_{net} = \sqrt{B_1^2 + B_2^2}$$

### The Final Calculation

Now, we simply substitute the magnitudes we found earlier into this equation:

$$B_{net} = \sqrt{\left(\frac{\mu_0 I_1}{2\pi d}\right)^2 + \left(\frac{\mu_0 I_2}{2\pi d}\right)^2}$$

Notice that the term $\left(\frac{\mu_0}{2\pi d}\right)^2$ is common to both terms inside the square root. We can factor it out:

$$B_{net} = \sqrt{\left(\frac{\mu_0}{2\pi d}\right)^2 (I_1^2 + I_2^2)}$$

Taking the square root of the common factor gives us our final, elegant expression for the net magnetic field at point P:

**$$B_{net} = \frac{\mu_0}{2\pi d}\sqrt{I_1^2 + I_2^2}$$**

This matches option (b), confirming our derivation is correct.

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Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$M a g n e t i c F i e l d o f an I n f ini t e W i r e$

$F i e l d s d u e t o I n d i v i d u a l W i r es$

$R es u l t an t M a g n e t i c F i e l d$

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$W ha t i f t h e p o in t w a s n o t o n t h e z - a x i s ?$

The Sigma Insight: Biot-Savart Law

Visualizing the Setup

The Magnetic Field of a Single Wire

Analyzing the Individual Fields

Vector Addition for the Net Field

The Final Calculation

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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 Setup

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fields due to Individual Wires

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the point was not on the z−axis?