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Physics Lesson 16.6.3 - Magnetic Field Outside a Long Straight Wire with Current

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Welcome to our Physics lesson on Magnetic Field Outside a Long Straight Wire with Current, this is the third lesson of our suite of physics lessons covering the topic of Ampere's Law, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Magnetic Field Outside a Long Straight Wire with Current

This situation is similar to the one discussed above. The only difference is that now we are considering a single current-carrying wire instead of two parallel or antiparallel ones. Therefore, the magnetic field B will have a cylindrical symmetry and any Ampere's loop we want to consider will have a circular shape, as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Ampere's Law

The current direction in this case is out of page. We enclose the current inside the desired Amperian loop of radius r, which is concentric to the wire's cross section. In this way, the integral (which is in the counter-clockwise direction) takes a simpler form than if the loop had a more complicated shape.

Giving that the magnetic field B⃗ is tangent to the loop, its vector is either parallel or antiparallel to the Amperian loop element dL⃗. For simplicity, it is better to assume them as parallel (θ = 0), so that cos θ = 1. Therefore, we obtain for Ampere's law in this situation:

∮B⃗ dL⃗ = ∮B⋅cos⁡θ dL
= B ∙ ∮dL
= B ∙ (2π ∙ r)
= μ0 ∙ iencl

Considering the last two identities

B ∙ (2π ∙ r) = μ0 ∙ iencl

we obtain the well-known formula of magnetic induction (in scalar form) produced by a long current carrying wire at distance r from the wire, which we have provided in tutorial 16.2:

B = μ0 ∙ iencl/2π ∙ r

Example 1

A magnetic field of magnitude 5mT is measured at a distance of 4cm from a long straight wire as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Ampere's Law

Calculate:

  1. The direction of electric current flowing through the wire
  2. The magnitude of this current

Solution 1

Clues:

r = 4 cm = 4 × 10-2 m
B = 0.5 mT = 5 × 10-4 T
(μ0 = 4π × 10-7 N/A2)

  1. The direction of current is found using the right hand rule. We grasp the wire by the right hand where the curled fingers show the direction of magnetic field. The outstretching thumb therefore shows the direction of current. In this case, this direction is from up to down.
  2. Using the equation obtained by applying Ampere's law for the magnetic field outside a long-straight wire,
    B = μ0 ∙ iencl/2π ∙ r
    we obtain for the electric current flowing through the wire
    iencl = B ∙ 2π ∙ r/μ0
    = (5 × 10-4 T) ∙ (2π) ∙ (4 × 10-2 m)/(4π × 10-7 N/A2 )
    = 102 A
    = 100A

You have reached the end of Physics lesson 16.6.3 Magnetic Field Outside a Long Straight Wire with Current. There are 5 lessons in this physics tutorial covering Ampere's Law, you can access all the lessons from this tutorial below.

More Ampere's Law Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
16.6Ampere's Law
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.6.1Useful Mathematical Background: Integral of a function - Geometrical approach
16.6.2Ampere's Law
16.6.3Magnetic Field Outside a Long Straight Wire with Current
16.6.4Magnetic Field Inside a Long Straight Wire with Current
16.6.5Ampere's Law Applied in Solenoids and Toroids

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