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CalibratePhysics

10.11

Force of Attraction between two Long Parallel Wires

Magnetic Fields due to Electric Current · Maharashtra HSC

Theory

Set-up and diagram

Diagram (Fig. 10.17): two long straight parallel wires, wire 1 and wire 2, drawn vertically in the plane of the paper a distance dd apart, carrying currents I1I_1 and I2I_2 in the same direction (both upward). The field B⃗\vec{B} produced by wire 1 at the position of wire 2 is marked as going into the plane of the paper (by the right hand thumb rule), and the force F⃗\vec{F} on wire 2 is drawn pointing towards wire 1.

This section applies the two results already obtained: the field of a long straight wire, and the force on a current carrying wire.

Field of wire 1 at wire 2

From the long-straight-wire result,

B=μ0I12πdB = \dfrac{\mu_0 I_1}{2\pi d}

By the right hand thumb rule, at the position of wire 2 this field is directed into the plane of the paper.

Force on wire 2

Applying the Lorentz force law in the form F=ILBF = I L B to a section of length LL of wire 2, which lies perpendicular to this field,

F21=I2BL=μ0I1I2L2πdF_{21} = I_2 B L = \dfrac{\mu_0 I_1 I_2 L}{2\pi d}

The direction of this force, from IL⃗×B⃗I\vec{L}\times\vec{B}, is towards wire 1: it is an attractive force when the currents are parallel.

For infinitely long wires the total force would be infinite, so the physically useful statement is the force per unit length:

FL=μ0I1I22πd\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}

Newton's third law

By the same argument, a section of length LL of wire 1 experiences a force F12F_{12} due to the current in wire 2, equal in magnitude and opposite in direction:

F⃗21=−F⃗12\vec{F}_{21} = -\vec{F}_{12}

This is consistent with Newton's third law, just as the gravitational pulls between the Earth and the Sun are equal and opposite.

Parallel and antiparallel currents

If the currents I1I_1 and I2I_2 flow in opposite directions, the direction of the force reverses and each wire is repelled from the other. The magnitude of the repulsive force per unit length is given by the same expression,

FL=μ0I1I22πd\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}

Summary: parallel currents attract, antiparallel currents repel. This settles the hanging-wires experiment of Section 10.1.

Definition of the ampere

Take two infinitely long parallel conducting wires placed in vacuum with a separation of 1 m. If the constant current through each produces a force of magnitude 2×10−72\times10^{-7} N per metre of length on the other, that current is 1 ampere.

Check: with μ04π=10−7\dfrac{\mu_0}{4\pi} = 10^{-7} Wb/m (that is, T m/A), d=1d = 1 m and I1=I2=1I_1 = I_2 = 1 A,

FL=μ0I1I22πd=2×10−7 N per metre\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} = 2\times10^{-7}\ \text{N per metre}

It is assumed here that the diameter of each wire is very much less than 1 m.

Standard results

  • B=μ0I12πdB = \dfrac{\mu_0 I_1}{2\pi d} at wire 2 due to wire 1
  • F=μ0I1I2L2πdF = \dfrac{\mu_0 I_1 I_2 L}{2\pi d} on a length LL; FL=μ0I1I22πd\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}
  • Parallel currents attract, antiparallel currents repel; F⃗21=−F⃗12\vec{F}_{21} = -\vec{F}_{12}
  • For equal currents, I=2πdμ0⋅FLI = \sqrt{\dfrac{2\pi d}{\mu_0}\cdot\dfrac{F}{L}}
  • Ampere: 2×10−72\times10^{-7} N/m between two infinite parallel wires 1 m apart in vacuum.

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

  1. 1

    [July 22] A current of equal magnitude flows through two long parallel wires separated by 2 cm. If force per unit length of 4×10−24\times10^{-2} N/m acts on both the wires respectively, calculate the current through each wire.

    Maharashtra HSC board questionAnswer 1 ↓
  2. 2

    [Feb 23] Calculate the current flowing through two long parallel wires carrying equal currents and separated by a distance of 1.35 cm experiencing a force per unit length of 4.76×10−24.76\times10^{-2} N/m.

    Maharashtra HSC board questionAnswer 2 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

  1. 1.↑ back to the question

    Given: I1=I2=II_1=I_2=I, FL=4×10−2\dfrac{F}{L}=4\times10^{-2} N/m, d = 2 cm = 2×10−22\times10^{-2} m To find: Electric current (I) Formula: FL=μ0I1I22πd\dfrac{F}{L}=\dfrac{\mu_0I_1I_2}{2\pi d} Calculation: From formula, 4×10−2=4π×10−7×I1×I22×π×2×10−24\times10^{-2}=\dfrac{4\pi\times10^{-7}\times I_1\times I_2}{2\times\pi\times2\times10^{-2}} ∴I2=4×10−210−5(∵I1=I2=I)\therefore I^2=\dfrac{4\times10^{-2}}{10^{-5}}\qquad(\because I_1=I_2=I) =4×10−2+5=4\times10^{-2+5} ∴I=4×103=4000=2010=63.24 A\therefore I=\sqrt{4\times10^3}=\sqrt{4000}=20\sqrt{10}=63.24\text{ A} Ans: The electric current through each wire is 63.24 A.

  2. 2.↑ back to the question

    Given: I1=I2=II_1=I_2=I, FL=4.76×10−2\dfrac{F}{L}=4.76\times10^{-2} N, d = 1.35 cm = 1.35×10−21.35\times10^{-2} m To find: Electric current Formula: FL=μ0I1I22πd\dfrac{F}{L}=\dfrac{\mu_0I_1I_2}{2\pi d} Calculation: From formula, 4.76×10−2=4π×10−7×I×I2×π×1.35×10−24.76\times10^{-2}=\dfrac{4\pi\times10^{-7}\times I\times I}{2\times\pi\times1.35\times10^{-2}} ∴I2=4.76×10−2×1.35×10−22×10−7\therefore I^2=\dfrac{4.76\times10^{-2}\times1.35\times10^{-2}}{2\times10^{-7}} =1.35×2.38×10−2−2+7=1.35\times2.38\times10^{-2-2+7} I=1.35×2.38×103=13.5×2.38×10=5.669×10=56.69 AI=\sqrt{1.35\times2.38\times10^3}=\sqrt{13.5\times2.38}\times10=5.669\times10=56.69\text{ A} Ans: The electric current is 56.69 A.