Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper.
The wire A

Q: Why is the magnetic field due to wire A at point P directed to the left?

Using the right-hand grip rule, point your thumb into the page (direction of current A). Your fingers will curl clockwise. At point P, which is directly below A, the tangent to this clockwise circle points to the left.

Q: How did we know the current in wire B is outwards?

To cancel the leftward magnetic field from wire A at point P, wire B must produce a rightward magnetic field. Pointing your fingers to the right at point P (which is below B) and curling them around B means your thumb must point out of the page.

Q: Why are the magnetic field vectors at S perpendicular to each other?

The magnetic field from a wire is always perpendicular to the radial line connecting the wire to the point. Since the radial lines AS and BS are perpendicular to each other (forming a right triangle), their corresponding magnetic fields are also perpendicular.

Q: Does the force between the wires depend on the point P or S?

No, the force per unit length between the two infinite parallel wires depends only on their currents and the perpendicular distance between them (which is 2 meters), independent of any other points in space.

Q: What would happen if the currents were in the same direction?

If the currents were in the same direction, their magnetic fields at point P would add up rather than cancel out, so the net field could never be zero at that location. Additionally, the force between the wires would be attractive instead of repulsive.

Question

Two long straight parallel wires are $2	ext{ m}$ apart, perpendicular to the plane of the paper.
The wire $A$ carries a current of $9.6	ext{ A}$, directed into the plane of the paper. The wire $B$ carries a current such that the magnetic field of induction at the point $P$, at a distance of $10/11	ext{ m}$ from the wire $B$, is zero.
Find
(a) the magnitude and direction of the current in $B$.
(b) the magnitude of the magnetic field of induction at the point $S$.
(c) the force per unit length on the wire $B$.

Sigma Percentile · Accepted Answer

### Decoding the Setup
Imagine two infinitely long parallel wires, $A$ and $B$, standing vertically like two pillars separated by a distance of $2	ext{ m}$. Wire $A$ carries a strong current of $9.6	ext{ A}$ plunging directly into the plane of the paper. Somewhere below wire $B$, at a specific point $P$ located $10/11	ext{ m}$ away, the universe perfectly balances itself—the net magnetic field is exactly zero. Our mission is to uncover the secrets of wire $B$ and explore the magnetic landscape around these wires.

### The Balancing Act at Point P
For the magnetic field to vanish at point $P$, the field produced by wire $A$ must be perfectly annihilated by the field from wire $B$. Let's apply the right-hand grip rule to wire $A$. With the thumb pointing into the page, our fingers curl clockwise. At point $P$, which lies directly below the wires, the tangent to this magnetic circle points sharply to the left.

To counter this leftward field, wire $B$ must generate a magnetic field pointing to the right at point $P$. Using the right-hand rule again, the only way wire $B$ can create a rightward field below it is if its current flows **outwards**, emerging from the plane of the paper.

Now, let's equate their magnitudes. The distance from $A$ to $P$ is $2 + 10/11 = 32/11	ext{ m}$.
$$B_A = B_B$$
$$\frac{\mu_0 i_A}{2\pi (32/11)} = \frac{\mu_0 i_B}{2\pi (10/11)}$$
Solving for $i_B$, we get:
$$i_B = i_A 	imes \frac{10}{32} = 9.6 	imes \frac{10}{32} = 3	ext{ A}$$

### The Geometry of Point S
Next, we shift our focus to point $S$. The distances given are $AS = 1.6	ext{ m}$, $BS = 1.2	ext{ m}$, and $AB = 2	ext{ m}$. Do these numbers look familiar? Let's check the squares:
$$1.6^2 + 1.2^2 = 2.56 + 1.44 = 4.0 = 2^2$$
This is a perfect Pythagorean triplet! Therefore, $	riangle ABS$ is a right-angled triangle, with the right angle firmly seated at $S$ ($\angle ASB = 90^\circ$).

Because the radial lines $AS$ and $BS$ are perpendicular, the magnetic field vectors $\vec{B}_A$ and $\vec{B}_B$ at point $S$ are also perpendicular to each other. Let's calculate their magnitudes:
$$B_A = \frac{\mu_0 i_A}{2\pi (1.6)} = \frac{2 	imes 10^{-7} 	imes 9.6}{1.6} = 12 	imes 10^{-7}	ext{ T}$$
$$B_B = \frac{\mu_0 i_B}{2\pi (1.2)} = \frac{2 	imes 10^{-7} 	imes 3}{1.2} = 5 	imes 10^{-7}	ext{ T}$$

Since they are orthogonal, the net magnetic field is simply the hypotenuse of this magnetic vector triangle:
$$B_{	ext{net}} = \sqrt{B_A^2 + B_B^2} = \sqrt{(12 	imes 10^{-7})^2 + (5 	imes 10^{-7})^2} = 13 	imes 10^{-7}	ext{ T}$$

### The Final Force
Finally, we evaluate the mechanical interaction between the two wires. Because the currents are anti-parallel (one into the page, one out of the page), they repel each other. The force per unit length is given by Ampere's force law:
$$\frac{F}{l} = \frac{\mu_0 i_A i_B}{2\pi d}$$
Substituting our known values:
$$\frac{F}{l} = \frac{4\pi 	imes 10^{-7} 	imes 9.6 	imes 3}{2\pi 	imes 2} = 2.88 	imes 10^{-6}	ext{ N/m}$$
This repulsive force constantly pushes wire $B$ away from wire $A$.

Visualized Solution

$S y s t e m S e t u p$

$M a g n e t i c F i e l d a tP d u e t o A$

$D i r ec t i o n o f C u r r e n t in B$

$M a g ni t u d eo f C u r r e n t in B$

$G eo m e t r y a tP o in tS$

$M a g n e t i c F i e l d s a tS$

$N e tM a g n e t i c F i e l d a tS$

$F or ce p er U ni t L e n g t h o n B$

The Sigma Insight: Ampere's Circuital Law

Decoding the Setup Imagine two infinitely long parallel wires, $A$ and $B$ , standing vertically like two pillars separated by a distance of $2 m$

The Balancing Act at Point P For the magnetic field to vanish at point $P$ , the field produced by wire $A$ must be perfectly annihilated by the field from wire $B$

The Geometry of Point S Next, we shift our focus to point $S$

The Final Force Finally, we evaluate the mechanical interaction between the two wires

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Visualized Solution

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