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# Two straight infinitely long and thin parallel wires are spaced $0.1{\text{m}}$ apart and carry a current of $10{\text{A}}$ each. Find the magnetic field at a point at a distance of $0.1{\text{m}}$ from both the wires in the two cases when the currents are in the: 1) Same direction, 2) Opposite direction

Last updated date: 18th Jun 2024
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Hint: The direction of the magnetic field at some point due to a current-carrying wire will always be perpendicular to the direction of the current in the wire. Here the magnitude of the magnetic field due to each parallel wire at the given point will be the same but will have different directions. So to obtain the net magnetic field at the given point we have to resolve the magnetic fields due to each wire into their respective components.

Formula used:
The magnetic field at a point due to an infinitely long current-carrying wire is given by, $B = \dfrac{{{\mu _0}I}}{{2\pi d}}$ where ${\mu _0}$ is the permeability of free space, $I$ is the current in the wire and $d$ is the distance between the wire and the point.

Step 1: Sketch a figure representing the arrangement of the two wires and the point at which the magnetic field is to be determined. Also, list the given parameters.

In the above figure, the two wires are considered to be placed so that they pass through the plane of the paper and points A and B represent any point of along the length of the two wires. P is the point at which the magnetic field due to the current in the two the wires is to be determined.
The current through each wire is given to be $I = 10{\text{A}}$ .
The distance between the two wires is given to be $AB = 0.1{\text{m}}$ .
The distance between point P and each wire is given to be $AP = AP = d = 0.1{\text{m}}$.
Since the triangle APB forms an equilateral triangle of side $0.1{\text{m}}$, the angle subtended at P will be $\angle P = 60^\circ$.

Step 2: Express the magnetic field at P due to each wire.
The magnetic field at point P due to wire A or wire B is equal in magnitude and so it can be expressed as ${B_A} = {B_B} = B = \dfrac{{{\mu _0}I}}{{2\pi d}}$ ---------- (1)
Substituting for $I = 10{\text{A}}$, $d = 0.1{\text{m}}$ and ${\mu _0} = 4\pi \times {10^{ - 7}}{\text{Tm}}{{\text{A}}^{ - 1}}$ in equation (1) we get, $B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 10}}{{2\pi \times 0.1}} = 2 \times {10^{ - 5}}{\text{T}}$
Thus the magnetic field at P due to each individual wire will be $B = 2 \times {10^{ - 5}}{\text{T}}$ .

Step 3: Take the direction of the current in the two wires to be the same and obtain the net magnetic field at P.
Case 1: Current is in the same direction in both wires
Let the direction of the current in both wires be out of the plane. Then by right-hand thumb rule, the direction of the magnetic field will be anticlockwise and at P the magnetic field due to each wire will have the directions given below.

If we resolve the magnetic fields at P due to each wire into horizontal and vertical components as shown in figure b, then only their horizontal components prevail.
So the net magnetic field at P will be ${B_{net}} = 2B\sin 60^\circ$ -------- (2)
Substituting for $B = 2 \times {10^{ - 5}}{\text{T}}$ and $\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$ in equation (2) we get, ${B_{net}} = 2 \times 2 \times {10^{ - 5}} \times \dfrac{{\sqrt 3 }}{2} = 2\sqrt 3 \times {10^{ - 5}}{\text{T}}$
$\therefore$ the net magnetic field at P when current is in the same direction in the two wires is ${B_{net}} = 2\sqrt 3 \times {10^{ - 5}}{\text{T}}$ .

Step 4: Take the direction of the current in the two wires to be the opposite and obtain the net magnetic field at P.
Case 1: Current is in the opposite direction in the two wires
Let the direction of the current in wire A be out of the plane and that in B be into the plane. Then by right-hand thumb rule, the direction of the magnetic field due to wire A will be anticlockwise while the direction of the magnetic field due to wire B will be clockwise. At P the magnetic field due to each wire will have the directions given below.

If we resolve the magnetic fields at P due to each wire into horizontal and vertical components as shown in figure d, then only their vertical components prevail.
So the net magnetic field at P will be ${B_{net}} = 2B\cos 60^\circ$ -------- (3)
Substituting for $B = 2 \times {10^{ - 5}}{\text{T}}$ and $\cos 60^\circ = \dfrac{1}{2}$ in equation (3) we get, ${B_{net}} = 2 \times 2 \times {10^{ - 5}} \times \dfrac{1}{2} = 2 \times {10^{ - 5}}{\text{T}}$

$\therefore$ The net magnetic field at P when current is in the opposite direction in the two wires is ${B_{net}} = 2 \times {10^{ - 5}}{\text{T}}$.

Note: According to the right-hand thumb rule, the right thumb indicates the direction of the current in the wire and the direction in which the remaining fingers curl around the thumb indicates the direction of the magnetic field. While resolving the components of the magnetic field at P, the vertical component will be the cosine component and the horizontal component will be the sine component. The net magnetic field in the first case will be along the negative x-direction and the net magnetic field in the second case will be along the positive y-direction.