Magnetic Field Due To Current In Straight Wire

Introduction

When an electric current flows through a straight wire, it generates a magnetic field around the wire. This phenomenon is a result of Ampère's law, which states that electric currents create magnetic fields. The magnetic field produced by a straight current-carrying wire forms concentric circles around the wire, and its direction is determined by the right-hand thumb rule: if you point your right thumb in the direction of the current, your fingers will curl in the direction of the magnetic field. The strength of the magnetic field decreases as the distance from the wire increases. This principle is foundational in understanding electromagnetism and is applied in devices such as electromagnets, transformers, and motors.

This Story also Contains

1. Introduction
2. Magnetic Field Due To Current In a Straight Wire:
3. Some Solved Examples
4. Summary

Magnetic Field Due To Current In a Straight Wire:

Magnetic field lines around a current-carrying straight wire are concentric circles whose centre lies on the wire.

The magnitude of magnetic field B, produced by a straight current-carrying wire at a given point is directly proportional to the current I pairing through the wire i.e. B is inversely proportional to the distance 'r' from the wire $\left ( B\propto \frac{1}{r} \right )$ as shown in the figure given below.

Derivation:

The directions of magnetic fields due to all current elements are the same in the figure shown, we can integrate the expression of magnitude as given by Biot-Savart law for the small current element dy as shown in the figure

B=∫dB=μ04π∫Idysin⁡θx2

In order to evaluate this integral in terms of angle $ivarphi$, we determine đy, x and ltheta in terms of perpendicular distance " r " (which is a constant ′′ϕ". Here,

y=rtan⁡ϕdy=rsec2⁡ϕdϕx=rsecϕ⁡ϕθ=π2−ϕ

Substituting in the integral, we have :

⇒B=μ04π∫Irsec2⁡ϕdϕsin⁡(π2−ϕ)r2sec2⁡ϕ=μ04π∫Icos⁡ϕdϕr

Taking out I and rout of the integral as they are constant:

⇒B=μ0I4πr∫cos⁡ϕdϕ

Integrating between angle −ϕ1 and ϕ2, we have

⇒B=μ0I4πr∫−ϕ1ϕ2Icos⁡ϕdϕ⇒B=μ0I4πr(sin⁡ϕ2−sin⁡(−ϕ1))

Note: −ϕ1 is taken because it is measured in the opposite sense of ϕ2 with respect to the reference line ( negative x-axis here)

⇒B=μ0I4πr(sin⁡ϕ2+sin⁡ϕ1)

Magnetic field due to a current-carrying wire at a point P which lies at a perpendicular distance r from the wire, as shown, is given as:

B=μ04πir(sin⁡ϕ1+sin⁡ϕ2)

From figure, α=(90∘−ϕ1) and β=(90∘+ϕ2)
Hence, it can be also written as B=μo4πir(cos⁡α−cos⁡β)

Different cases:

Case 1: When the linear conductor XY is of finite length and the point P lies on it's perpendicular bisector as shown

B=μ04π⋅ir(2sin⁡ϕ)

Case 2: When the linear conductor XY is of infinite length and the point P lies near the centre of the conductor

ϕ1=ϕ2=90∘.B=μ04πir[sin⁡90∘+sin⁡90∘]=μ04π2ir

Case 3: When the linear conductor is of semi-infinite length and the point P lies near the end Y or X

ϕ1=90∘ and ϕ2=0∘B=μ04πir[sin⁡90∘+sin⁡0∘]=μ04πir

Case 4: When point P lies on the axial position of the current-carrying conductor then magnetic field at P,
- P
α=β=0∘

B=μo4πir(cos⁡α−cos⁡β)=μo4πir(cos⁡0−cos⁡0)=0

Note:

• The value of magnetic field induction at a point, on the centre of separation of two linear parallel conductors carrying equal currents in the same direction, is zero.
• If the direction of current in the straight wire the known then the direction of the magnetic field produced by a straight wire carrying current is obtained by maxwell's right-hand thumb rule.

Some Solved Examples

Example 1: A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5×10−2 m. The magnetic field (in Wb/m2 ) at the centre of the triangle will be :

1) 4×10−5 2) 0 3) 8×10−5 4) 2×10−5

Solution:

B→BC=μ0I4πa23(sin⁡60∘+sin⁡60∘)=μ0I2πa3(3)=3μ0I2πa=3∗2∗10−7∗14.5∗10−2=21.5∗10−5 Wb/m2

∴ Total Magnetic field

=3∗2∗10−51.5=4∗10−5 Wb/m2

Example 2: The magnetic field at the origin due to the current flowing in the wire as shown in the figure below is

1) −μ0I8πa(i^+k^) 2) μ0I2πa(i^+k^) 3) μ0I8πa(−i^+k^)μ0I8πa(i^−k^)

Solution:
Magnetic Field Due to a Straight Wire -

B=μoi4πr(sin⁡ϕ1+sin⁡ϕ2)

OR

B=μo4πir(cos⁡α−cos⁡β)

Here, the wire along the x -axis and the wire in the x -z plane pass through the origin.

So the magnetic field at the origin is zero.

Now for the wire parallel to the y-axis

ϕ2=0 and ϕ1=90∘

So

B0=μ0I4πa2

and its direction is along

B^o=dl^×r^=j^×(−i^+−k^)2=cos⁡45∘(^−i)+cos⁡45∘k^

So

B0=μ0I4πa2[cos⁡45∘(^−i)+cos⁡45∘k^)]=μ0I8πa(−i^+k^)

Example 2: Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (see figure) (μ0=4π×10−7N−A−2)

1) 2.0×10−5 T
2) 1.5×10−5 T
3) 3.0×10−5 T
4) 2.5×10−5 T

Solution

sin⁡Θ=35 So Bp=μ0I4πd×2sin⁡ΘBp=107×4π4π×5×24×10−2×35Bp=1.5×10−5T

Summary

A current-carrying straight wire generates a magnetic field that forms circular loops around the wire. The direction of the magnetic field is given by the right-hand rule, with the magnetic field lines encircling the wire. The strength of the magnetic field depends on the current and the distance from the wire, decreasing as you move further away from the wire. This concept is key in explaining the interaction between electricity and magnetism, forming the basis of many practical applications in electrical engineering, such as in solenoids, inductors, and other devices that use magnetic fields generated by current-carrying wires.