Force between Two Parallel Currents, the Ampere

Moving Charges and Magnetism · CBSE

Theory

Setting up the derivation

Two long straight parallel conductors $a$ and $b$ , separated by a distance $d$ , carry steady currents $I_a$ and $I_b$ in the same direction.
Conductor $a$ produces, at every point along conductor $b$ , the same magnetic field $B_a = \dfrac{\mu_0 I_a}{2\pi d}$ directed perpendicular to the plane of the two wires (by the right-hand thumb rule; if the wires are horizontal and the currents point away from you, this field is directed downwards at $b$ , i.e. into the plane containing the wires).
Conductor $b$ , carrying $I_b$ , then feels a sideways force in this field. On a segment of length $L$ , using $\vec{F} = I(\vec{L}\times\vec{B})$ with $\vec{L}\perp\vec{B}_a$ : $F_{ba} = I_b L B_a = \dfrac{\mu_0 I_a I_b}{2\pi d}\,L$
The direction of $\vec{L}\times\vec{B}_a$ is towards conductor $a$ : the wires attract.
Repeating the argument for the force on $a$ due to $b$ gives an equal magnitude directed towards $b$ , so $\vec{F}_{ba} = -\vec{F}_{ab}$ consistent with Newton's third law for steady currents.
Force per unit length: $f = \dfrac{F}{L} = \dfrac{\mu_0 I_a I_b}{2\pi d} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2 I_a I_b}{d} \quad \left(\text{N m}^{-1}\right)$

The rule, and how it differs from electrostatics

Parallel currents attract; antiparallel currents repel.
This is the opposite of the electrostatic rule: like charges repel, but like (parallel) currents attract. CBSE asks for the contrast explicitly.
Reversing one current reverses the field it produces at the other wire, so the force on each wire reverses: the wire on the left is pushed further left and the one on the right further right — they repel.

Definition of the ampere

Put $I_a = I_b = 1\ \text{A}$ and $d = 1\ \text{m}$ : $f = \dfrac{\mu_0}{2\pi} = 2\times10^{-7}\ \text{N m}^{-1}$
The ampere is the value of that steady current which, when maintained in each of two very long, straight, parallel conductors of negligible cross-section placed one metre apart in vacuum, produces on each of these conductors a force of $2\times10^{-7}$ newton per metre of length.
This definition (adopted in 1946) is theoretical; in practice the earth's field must be eliminated and multiturn coils used, with a current balance measuring the force.
The coulomb follows: when a steady current of $1\ \text{A}$ is set up in a conductor, the charge crossing a section in one second is one coulomb.

Diagram in words

Two long vertical (or horizontal) parallel lines labelled $a$ and $b$ , with arrows for $I_a$ and $I_b$ in the same direction, and $d$ marked between them.
At wire $b$ , draw the field $\vec{B}_a$ as a cross (into the page) and the force $\vec{F}_{ba}$ as an arrow pointing towards $a$ ; mirror it at wire $a$ . For the antiparallel case, redraw with the two force arrows pointing outwards.

Loop near a straight wire

A square or rectangular loop of side $L$ carrying $I_2$ lies in the same plane as a long straight wire carrying $I_1$ , with the near side at distance $r_1$ and the far side at $r_2$ .
The two sides perpendicular to the wire carry current in opposite senses at each distance and their forces cancel by symmetry.
The near side (current parallel to $I_1$ ) is attracted with $F_{\text{near}} = \dfrac{\mu_0 I_1 I_2 L}{2\pi r_1}$ ; the far side (current antiparallel) is repelled with $F_{\text{far}} = \dfrac{\mu_0 I_1 I_2 L}{2\pi r_2}$ . Net: $F = \dfrac{\mu_0 I_1 I_2 L}{2\pi}\left[\dfrac{1}{r_1}-\dfrac{1}{r_2}\right] \quad\text{towards the wire}$
Square loop of side $L$ with the near side at $L$ and far side at $2L$ : $F_{\text{near}} = \dfrac{\mu_0 I_1 I_2}{2\pi}$ , $F_{\text{far}} = \dfrac{\mu_0 I_1 I_2}{4\pi}$ , net $F = \dfrac{\mu_0 I_1 I_2}{4\pi}$ — which is exactly equal to the force on the far side. So "the force on the far side" equals the net force $F$ on the loop.
Numerical: square loop of side $20\ \text{cm}$ carrying $1\ \text{A}$ , near side $10\ \text{cm}$ and far side $30\ \text{cm}$ from a wire carrying $2\ \text{A}$ : $F = \dfrac{\mu_0}{\pi}\left[2-\dfrac{2}{3}\right] = \dfrac{4\mu_0}{3\pi} = 5.33\times10^{-7}\ \text{N}$ , attractive.

Equilibrium of a third wire

Wires $A_1$ and $A_2$ carrying $I$ and $2I$ in the same direction are $d$ apart. A third wire $A_3$ carrying $1.5I$ is placed between them at distance $x$ from $A_1$ ; the forces from the two sides are attractive and oppositely directed. Setting them equal: $\dfrac{\mu_0 (I)(1.5I)}{2\pi x} = \dfrac{\mu_0 (2I)(1.5I)}{2\pi (d-x)} \;\Rightarrow\; \dfrac{1}{x} = \dfrac{2}{d-x} \;\Rightarrow\; x = \dfrac{d}{3}$ so $A_3$ sits $d/3$ from $A_1$ and $2d/3$ from $A_2$ .
The current in $A_3$ cancels from the equation, so the equilibrium *position* does not depend on it — though the magnitudes of the individual forces do.
Three wires $A$ , $B$ , $C$ carrying $2I$ , $I$ , $I$ in a row at equal spacing $r$ , all initially in the same sense: the forces on $C$ from $A$ (distance $2r$ ) and from $B$ (distance $r$ ) are each $\dfrac{\mu_0 I^{2}}{2\pi r}$ and add, so $F = \dfrac{2\mu_0 I^{2}}{2\pi r}$ . Reversing the current in $A$ leaves the magnitudes unchanged but makes that force repulsive, so the two now cancel and the net force on $C$ is zero.

Force on a moving charge near a current-carrying wire

The wire's field at the charge is $B = \dfrac{\mu_0 I}{2\pi y}$ ; the force on a charge $q$ moving with speed $v$ perpendicular to that field is $F = qvB = \dfrac{\mu_0 I q v}{2\pi y}$ with the direction from $\vec{F} = q\vec{v}\times\vec{B}$ (or Fleming's left-hand rule for a positive charge). A positive charge moving parallel to the current is attracted towards the wire; moving antiparallel, it is repelled.

Worked numericals

Two wires $2\ \text{m}$ apart, $3\ \text{A}$ each, same direction: $\dfrac{F}{L} = \dfrac{\mu_0}{2\pi}\cdot\dfrac{3\times3}{2} = 2\times10^{-7}\times4.5 = 9\times10^{-7}\ \text{N m}^{-1}$ , attractive.
Wires $2.5\ \text{cm}$ apart carrying $4\ \text{A}$ and $10\ \text{A}$ : $\dfrac{F}{L} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2 I_1 I_2}{r} = \dfrac{10^{-7}\times2\times4\times10}{2.5\times10^{-2}} = 3.2\times10^{-4}\ \text{N m}^{-1}$ .

Results to carry away

$B_a = \dfrac{\mu_0 I_a}{2\pi d}$ , $F = \dfrac{\mu_0 I_a I_b L}{2\pi d}$ , $f = \dfrac{\mu_0 I_a I_b}{2\pi d}$ .
Parallel attract, antiparallel repel; $\vec{F}_{ba} = -\vec{F}_{ab}$ .
Ampere: $2\times10^{-7}\ \text{N m}^{-1}$ at $1\ \text{m}$ for $1\ \text{A}$ in each.
Loop near a wire: perpendicular sides cancel, net force $\propto\left[\dfrac{1}{r_1}-\dfrac{1}{r_2}\right]$ .

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
A square shaped current carrying loop $MNOP$ (side $L$ , carrying current $I_2$ ) is placed near a straight long current carrying wire $AB$ (carrying current $I_1$ ), in the same plane, with the nearer side $PM$ at distance $L$ from the wire and the farther, parallel side $NO$ at distance $2L$ from the wire (the top and bottom sides $MN$ and $OP$ being perpendicular to the wire, spanning from distance $L$ to $2L$ ). If the loop experiences a net force $F$ towards the wire, find the magnitude of the force on the side ' $NO$ ' of the loop.
1 markCBSE Class 12 Physics — Very Short Answer · 2020, Ap · Section A, 1 markAnswer 1 ↓
2
Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current.
2 marksCBSE Class 12 Physics — SA · AI 2014 · Section B, 2 marksAnswer 2 ↓
3
A long straight conductor kept along $X'X$ axis carries a steady current $I$ along $+x$ direction. At an instant $t$ , a particle of mass $m$ and charge $q$ at point $(x,y)$ moves with a velocity $\vec{v}$ along $+y$ direction. Find the magnitude and direction of the force on the particle due to the conductor.
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 3 ↓
4
A square loop $ABCD$ of side 20 cm carrying current of 1 A is kept near an infinite long straight wire carrying a current of 2 A in the same plane, with the nearer side $AB$ at 10 cm from the wire and the farther side $CD$ at 30 cm from the wire. Calculate the magnitude and direction of the net force exerted on the loop due to the current carrying conductor.
2 marksCBSE Class 12 Physics — SA · AI 2015C, Ap · Section B, 2 marksAnswer 4 ↓
5
Two infinitely long straight wires $A_1$ and $A_2$ carrying currents $I$ and $2I$ flowing in the same directions are kept 'd' distance apart. Where should a third straight wire $A_3$ carrying current $1.5I$ be placed between $A_1$ and $A_2$ so that it experiences no net force due to $A_1$ and $A_2$ ? Does the net force acting on $A_3$ depend on the current flowing through it?
3 marksCBSE Class 12 Physics — SA · Delhi 2019 · Section C, 3 marksAnswer 5 ↓
6
Two long straight parallel conductors carry steady current $I_1$ and $I_2$ separated by a distance $d$ . If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.
3 marksCBSE Class 12 Physics — SA · Delhi 2016, An · Section C, 3 marksAnswer 6 ↓
7
(a) Two long straight parallel conductors 'a' and 'b' carrying steady currents $I_a$ and $I_b$ are separated by a distance $d$ . Write the magnitude and direction of the magnetic field produced by the conductor ' $a$ ' at the points along the conductor ' $b$ '. If the currents are flowing in the same direction, what is the nature and magnitude of the force between the two conductors? (b) Show with the help of a diagram how the force between the two conductors would change when the currents in them flow in the opposite directions.
3 marksCBSE Class 12 Physics — SA · Foreign 2014 · Section C, 3 marksAnswer 7 ↓

Numericals

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

8
Two long parallel wires kept 2 m apart carry 3 A current each, in opposite directions. The force per unit length on one wire due to the other is
- (a)
  $4.5\times10^{-5}$ Nm $^{-1}$ , attractive
- (b)
  $4.5\times10^{-7}$ N/m, repulsive
- (c)
  $9\times10^{-7}$ N/m, repulsive
- (d)
  $9\times10^{-5}$ N/m, attractive
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 8 ↓
9
Two parallel wires carrying currents $I_1$ and $I_2$ (in opposite directions) lie, one slightly above the other, in a horizontal plane, dividing the surrounding region into four quadrant-like zones I, II, III and IV (I and II above the wires, III and IV below, split left/right by the wires' relative offset). The region of vertically upward strongest magnetic field is
- (a)
  I
- (b)
  II
- (c)
  III
- (d)
  IV
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 9 ↓
10
Two parallel conductors carrying current of 4.0 A and 10.0 A are placed 2.5 cm apart in vacuum. The force per unit length between them is
- (a)
  $6.4\times10^{-5}$ N m $^{-1}$
- (b)
  $6.4\times10^{-2}$ N m $^{-1}$
- (c)
  $4.6\times10^{-4}$ N m $^{-1}$
- (d)
  $3.2\times10^{-4}$ N m $^{-1}$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 10 ↓

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.↑ back to the question
Force between two parallel current carrying conductors, $F = \dfrac{\mu_0 I_1 I_2 L}{2\pi r}$ Force on side $PM$ (nearer to wire $AB$ , at distance $L$ ): $F_{PM} = \dfrac{\mu_0 I_1 I_2 L}{2\pi L} = \dfrac{\mu_0 I_1 I_2}{2\pi}$ (attractive, towards the wire). Force on side $NO$ (farther, at distance $2L$ ): $F_{NO} = \dfrac{\mu_0 I_1 I_2 L}{2\pi(2L)} = \dfrac{\mu_0 I_1 I_2}{4\pi}$ (repulsive, away from the wire). The forces on sides $MN$ and $OP$ (perpendicular to the wire) are equal and opposite (by symmetry, since the current flows in opposite senses along them), so they cancel out. Net force on the loop, towards the wire: $F = F_{PM} - F_{NO} = \dfrac{\mu_0 I_1 I_2}{2\pi} - \dfrac{\mu_0 I_1 I_2}{4\pi} = \dfrac{\mu_0 I_1 I_2}{4\pi}$ Comparing with $F_{NO}$ above, we see $F_{NO} = \dfrac{\mu_0 I_1 I_2}{4\pi} = F$ . So, the magnitude of the force on the side $NO$ of the loop equals $F$ , the same as the net force on the loop.
2.↑ back to the question
Force per unit length between two parallel current carrying conductors: $F = \dfrac{\mu_0 I_1 I_2}{2\pi r}$ When $I_1 = I_2 = 1$ A and $r = 1$ m, then $F = 2\times10^{-7}$ N m $^{-1}$ . One ampere is the value of steady current which, when maintained in each of two very long, straight, parallel conductors of negligible cross-section and placed one metre apart in vacuum, would produce on each of these conductors a force of attractive or repulsive nature of magnitude $2\times10^{-7}$ N per metre of their length.
3.↑ back to the question
Magnetic field due to the wire at point $(x,y)$ (distance $y$ from the wire): $B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I}{y} = \dfrac{\mu_0 I}{2\pi y}$ , directed out of the plane of paper (by right hand thumb rule, for current along $+x$ and the point above the wire). Force on charge $q$ moving with velocity $v$ (along $+y$ ), perpendicular to $B$ (which is along $+z$ ): $F = qvB\sin90° = qv\cdot\dfrac{\mu_0 I}{2\pi y}$ $F = \dfrac{\mu_0 Iqv}{2\pi y}$ Direction is given by Fleming's left hand rule (or $\vec{F} = q\vec{v}\times\vec{B}$ ), and it is found to be towards the '+x' axis (i.e. towards the wire, along the direction of the current in the wire).
4.↑ back to the question
Force on arm $AB$ (nearer side of the square loop, at distance 10 cm from the wire): $F_{AB} = \dfrac{\mu_0\times2\times1\times20\times10^{-2}}{2\pi\times10\times10^{-2}} = \dfrac{2\mu_0}{\pi}$ N (attractive, towards the wire) Force on arm $CD$ (farther side, at distance $10+20 = 30$ cm from the wire): $F_{CD} = \dfrac{\mu_0\times2\times1\times20\times10^{-2}}{2\pi\times30\times10^{-2}} = \dfrac{2\mu_0}{3\pi}$ N (repulsive, away from the wire) Force on arms $BC$ and $DA$ are equal and opposite (both perpendicular to the wire, with equal and opposite net contributions along the wire's direction), so they cancel out each other. Net force on the loop is $F = F_{AB} - F_{CD} = \dfrac{\mu_0}{\pi}\left[2 - \dfrac{2}{3}\right] = \dfrac{4\mu_0}{3\pi} = \dfrac{4\times4\pi\times10^{-7}}{3\pi}$ $F = 5.33\times10^{-7}$ N, attractive (directed towards the wire)
5.↑ back to the question
Force on $A_3$ due to $A_1$ (at distance $x$ ): $f_1 = \dfrac{4\pi\times10^{-7}\times I\times1.5I}{2\pi x}$ Force on $A_3$ due to $A_2$ (at distance $d-x$ ): $f_2 = \dfrac{4\pi\times10^{-7}\times2I\times1.5I}{2\pi(d-x)}$ When there is no net force on $A_3$ : $f_1 = f_2$ $\dfrac{4\pi\times10^{-7}\times I\times1.5I}{2\pi x} = \dfrac{4\pi\times10^{-7}\times2I\times1.5I}{2\pi(d-x)}$ $\Rightarrow \dfrac{1}{x} = \dfrac{2}{d-x} \Rightarrow d-x = 2x \Rightarrow x = \dfrac{d}{3}$ Hence, wire $A_3$ should be placed at a distance $d/3$ from $A_1$ (i.e. $2d/3$ from $A_2$ ). Also, from the above result, we can say that the net (equilibrium) position is independent of the current flowing in $A_3$ — the current $1.5I$ cancelled out of the equation, so the position (though not the magnitude of the individual forces) does not depend on the current flowing through $A_3$ .
6.↑ back to the question
When two parallel infinite straight wires carrying currents $I_1$ and $I_2$ are placed at distance $d$ from each other, then current $I_1$ produces magnetic field, which at any point on the second current carrying wire is $B_1 = \dfrac{\mu_0 I_1}{2\pi d}$ , directed inwards, perpendicular to the plane of wires (by the right hand rule, for the given current directions). So, this current-carrying wire (carrying $I_2$ ) then experiences a force due to this magnetic field which on its length $l$ is given by $\vec{F}_{21} = I_2(\vec{l}\times\vec{B}_1)$ $F_{21} = F_{12} = I_2 l B_1\sin90° = I_2 l\times\dfrac{\mu_0 I_1}{2\pi d}$ or $F_{21} = F_{12} = \dfrac{\mu_0 I_1 I_2}{2\pi d}l$ The vector product $(\vec{l}\times\vec{B}_1)$ has a direction towards the wire carrying current $I_1$ . Hence, both the wires attract each other. So, force per unit length that each wire exerts on the other is $F = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ If $I_1 = I_2 = 1$ A and $d = 1$ m and $l = 1$ m, then $F = \dfrac{\mu_0}{2\pi} = 2\times10^{-7}$ N m $^{-1}$ Thus, electric current through each of two parallel long wires placed at distance of 1 m from each other is said to be 1 ampere, if they exert a force of $2\times10^{-7}$ N m $^{-1}$ on each other.
7.↑ back to the question
(a) When two parallel infinite straight wires carrying currents $I_a$ and $I_b$ are placed at distance $d$ from each other, then current $I_a$ produces magnetic field, which at any point on the second wire $b$ is $B_1 = \dfrac{\mu_0 I_a}{2\pi d}$ , directed inwards, perpendicular to the plane of wires. This current ( $I_b$ ) carrying wire then experiences a force per unit length $F = \dfrac{\mu_0 I_a I_b}{2\pi d}$ , and since the currents flow in the same direction, this force is attractive in nature. (b) Now, let the direction of current $b$ be reversed. The magnetic field $B_2$ at point $P$ due to current $I_a$ flowing through $a$ will be downwards. Similarly, the magnetic field $B_1$ at point $Q$ due to current $I_b$ passing through $b$ will also be downwards. The force on $a$ will be, therefore, towards the left. Also, the force on $b$ will be towards the right. Hence, the two conductors will repel each other when the currents flow in opposite directions.
8.Answer: (C)↑ back to the question
Force per unit length between two long parallel wires: $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ With $I_1=I_2=3$ A and $d=2$ m: $\dfrac{F}{L} = \dfrac{4\pi\times10^{-7}}{2\pi}\times\dfrac{3\times3}{2} = 2\times10^{-7}\times4.5 = 9\times10^{-7}\text{ N/m}$ The currents are in opposite directions, and anti-parallel currents REPEL. So the force per unit length is $9\times10^{-7}$ N/m, repulsive -- option (c).
[Note on this record: the stem was transcribed as "in the same direction", which no option answers, since same-direction currents attract and both $10^{-7}$ options say repulsive. The option set pairs each magnitude with each direction, placing the correct magnitude $9\times10^{-7}$ on "repulsive", so the question was set for anti-parallel currents. The stem is corrected; the printed key was not changed.]
9.Answer: (B)↑ back to the question
Answer: (b) II The magnetic field due to each wire is strongest closest to that wire, and its direction (up or down) depends on the current direction (right-hand thumb rule) and on which side of the wire the point lies. Analysing the direction of the field due to each wire in each of the four zones, the fields due to the two (oppositely-directed) currents add constructively (both pointing vertically upward) only in region II — in the other three regions the two contributions either partially cancel or do not both point upward.
10.Answer: (D)↑ back to the question
Answer: (d) $3.2\times10^{-4}$ N m $^{-1}$ $I_1 = 4$ A, $I_2 = 10$ A, $r = 2.5$ cm $= 2.5\times10^{-2}$ m Force per unit length: $F = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I_1I_2}{r} = \dfrac{10^{-7}\times2\times4\times10}{2.5\times10^{-2}}$ $F = 3.2\times10^{-4}$ N m $^{-1}$