Magnetic Force on Current Carrying Wires

Q: What is the force on a current-carrying wire in a magnetic field?

For a straight wire of length L carrying current I in uniform B, F = I(L×B), magnitude F = BIL sin θ. Direction: right-hand rule for I×B, or Fleming's left-hand rule.

Q: Why is the net force on a closed loop zero in a uniform field?

Net force on any wire in a uniform field equals I(L_endpoints × B). For a closed loop, start and end coincide, so L_endpoints = 0 and net force vanishes. A loop can still feel a torque.

Q: What is the force per unit length between two parallel current-carrying wires?

F/L = μ₀ I₁ I₂/(2πd). Same-direction currents attract; opposite-direction currents repel.

Q: Why do parallel currents attract each other?

Wire 1's magnetic field at wire 2 is oriented so that the force I₂ L₂ × B₁ pulls wire 2 toward wire 1. Newton's third law gives the reverse for wire 1.

Q: How is the SI ampere defined?

Historically: one ampere is the current that gives 2×10⁻⁷ N/m between two parallel wires 1 m apart. Post-2019, the ampere is defined via the fixed elementary charge e.

Q: Does the shape of a wire between two endpoints affect the total force in a uniform field?

No. Only the straight vector from start to end matters: F = I(L_endpoints × B). Any shape with the same endpoints feels the same total force.

Q: How can a current-carrying wire be levitated by another current?

Place a horizontal wire above a parallel wire, currents in opposite directions. The repulsive (upward) magnetic force per unit length balances gravity at height d = μ₀I₁I₂/(2πmg).

Q: What is Fleming's left-hand rule?

With thumb, index, and middle finger of left hand mutually perpendicular: index = B, middle = I, thumb = F on the wire. Equivalent to the right-hand rule for I×B.

Q: Why does a rectangular loop near a long straight wire feel a net force despite the closed-loop rule?

The closed-loop-zero-force rule holds only in uniform fields. Near a straight wire, the field varies as 1/r, so the near and far sides feel different forces, leaving a net attraction or repulsion.

PhysicsMagnetic Effects of Current and MagnetismFor NEET aspirants

A current-carrying wire in a magnetic field feels a force because the moving charges inside it feel the Lorentz force. For a small element $d ℓ$ carrying current $I$ , the force is $d F = I (d ℓ \times B)$ . Integrating: in a uniform field, the total force on any wire depends only on the straight-line vector joining its endpoints, $F = I (L \times B)$ - which means a closed loop feels zero net force in a uniform field. The other tested consequences are the force per unit length between two parallel wires, $F / L = μ_{0} I_{1} I_{2} / (2 π d)$ , and the SI definition of the ampere that follows from it.

Key Formulas - Quick Reference

Force on a current element: $d F = I d ℓ \times B$
Straight wire of length $L$ in uniform $B$ : $F = I (L \times B)$ , magnitude $F = B I L sin θ$
Arbitrary shape in uniform $B$ : $F = I (L_{endpoints} \times B)$ , where $L_{endpoints}$ is the straight vector from start to end
Closed loop in uniform $B$ : $F_{net} = 0$
Force per unit length between parallel wires (separation $d$ ): $\frac{F}{L} = \frac{μ _{0} I _{1} I _{2}}{2 π d}$
Parallel currents attract; antiparallel currents repel
Definition of ampere: current giving $2 \times 1 0^{- 7}$ N/m between two long parallel wires $1$ m apart

Colours in figures: magnetic field

B

current

I

force

F

1Force on a Current Element

Inside a wire, free charges (electrons) drift with speed $v_{d}$ . The current is $I = n e v_{d} A$ , where $n$ is number density and $A$ is cross-section. Each moving charge feels a Lorentz force. Summing over all $n A d ℓ$ charges in an element of length $d ℓ$ :

d F = I d ℓ \times B

Here $d ℓ$ points in the direction of conventional current $I$ . The direction of $d F$ follows the right-hand rule for the cross product.

Figure 1(a) A straight wire carrying current

I

to the right in a uniform field into the page feels

F = I L \times B

upward, spread evenly along its length. (b) Fleming's left-hand rule: first finger along

B

, second finger along

I

, thumb gives

F

. The three are mutually perpendicular and match panel (a).

Directional rules

Right-hand rule for $I \times B$ : fingers point along $I$ , curl toward $B$ ; thumb gives $F$ .
Fleming's left-hand rule: stretch thumb, index, middle finger mutually perpendicular. Index = field $B$ ; middle = current $I$ ; thumb = force $F$ .
Both give the same answer; use whichever you're comfortable with.

2Straight Wire in a Uniform Field

For a straight wire of length $L$ in uniform $B$ , integration gives:

F = I (L \times B) F = B I L sin θ

where $L$ is a vector of length equal to the wire, pointing along the current, and $θ$ is the angle between $L$ and $B$ . Force is zero when the wire is parallel to $B$ , maximum when perpendicular.

3Arbitrary Shape in a Uniform Field

For a curved or bent wire in uniform $B$ , the total force is:

$F = I \int d ℓ \times B = I (\int d ℓ) \times B = I L_{endpoints} \times B$

where $L_{endpoints}$ is the straight-line vector from the start to the end of the wire, not the arc length. The wire's shape between the endpoints does not matter for the force in a uniform field!

Figure 2In a uniform field the force on a bent wire equals the force on the straight wire joining its ends:

F = I L_{endpoints} \times B

. Every small piece

d ℓ

contributes

I d ℓ \times B

, and adding all the

d ℓ

vectors gives the chord

L

from

A

C

Key corollary: for a closed loop (any shape) in a uniform field,

L_{endpoints} = 0

, so

F_{net} = 0

. A closed loop can still feel a torque in a uniform field, but never a net force. Net force requires a non-uniform field (Concept 261 covers this).

Figure 3Closed loops in a uniform field (into the page, anticlockwise current). Left: forces on opposite sides of a rectangle are equal and opposite. Right (Solved Example 2): each side of the triangle feels

F = B I L

perpendicular to it; the three forces add to zero. In a uniform field, any closed loop feels zero net force.

Solved Example 1

A wire bent into a semicircle of radius

R

lies in the

x y

-plane, carrying current

I

from one end of the diameter to the other. A uniform field

B = B_{0} \hat{k}

(out of the plane) exists. Find the net force on the semicircle.

Figure 4Solved Example 1: a semicircular wire from

P

Q

in a uniform field out of the page. The force equals that on the diameter

P Q

carrying the same current:

F = I (2 R) B_{0}

, perpendicular to

P Q

, along

- \hat{j}

Solution:

Instead of integrating the arc, use the shortcut: $L_{endpoints}$ is the straight diameter, length $2 R$ along, say, the $x$ -axis.

$F = I (L \times B) = I (2 R \hat{i}) \times (B_{0} \hat{k}) = 2 I R B_{0} (\hat{i} \times \hat{k}) = - 2 I R B_{0} \hat{j}$

Magnitude $2 I R B_{0}$ , direction along $- \hat{j}$ (perpendicular to the diameter, in the plane).

Solved Example 2

A wire is bent into an equilateral triangle PQR of side

10

cm and carries

5.0

A. It is placed in a uniform

B = 2.0

T perpendicular to the plane of the loop. Find the force on each of the three sides.

Solution:

Each side has length $0.10$ m and carries the current perpendicular to $B$ , so on each side:

$F = B I L = 2.0 \times 5.0 \times 0.10 = 1.0$ N.

Direction: perpendicular to that side, in the plane of the triangle, pointing inward for anticlockwise current with $B$ into the page (reversing either one makes all three point outward). The three forces of $1.0$ N form a symmetric set that sums to zero, consistent with the closed-loop result.

4Force Between Two Parallel Wires

Consider two long parallel wires carrying currents $I_{1}$ and $I_{2}$ , separated by distance $d$ . Wire 1 creates a field $B_{1} = \frac{μ _{0} I _{1}}{2 π d}$ at the position of wire 2. The force per unit length on wire 2:

\frac{F}{L} = I_{2} B_{1} = \frac{μ _{0} I _{1} I _{2}}{2 π d}

Figure 5Two long parallel wires seen end-on. Wire 1's field

B_{1} = \frac{μ _{0} I _{1}}{2 π d}

at wire 2 acts on

I_{2}

. (a) Currents in the same direction:

F

on each wire points towards the other, so they attract. (b) Opposite directions: the forces reverse and the wires repel. In both cases

\frac{F}{L} = \frac{μ _{0} I _{1} I _{2}}{2 π d}

, equal and opposite on the two wires.

Same direction (parallel currents): forces are attractive.
Opposite directions (antiparallel currents): forces are repulsive.
Both wires experience equal and opposite forces (Newton's third law).

5SI Definition of the Ampere

Setting $I_{1} = I_{2} = 1$ A and $d = 1$ m in the parallel-wire formula:

$\frac{F}{L} = \frac{μ _{0} \cdot 1 \cdot 1}{2 π \cdot 1} = 2 \times 1 0^{- 7} N/m$

Historical SI definition (pre-2019): One ampere is the constant current that, flowing in each of two infinitely long, thin parallel wires placed

1

m apart in vacuum, produces a force of

2 \times 1 0^{- 7}

N per metre length between them.

Post-2019 SI: the ampere is now defined via the elementary charge

e = 1.602176634 \times 1 0^{- 19}

C (exact), and

μ_{0}

is a measured constant extremely close to but no longer exactly

4 π \times 1 0^{- 7}

T·m/A. The old definition remains standard in JEE and NEET syllabi and gives the same numerical relations for all practical purposes.

Solved Example 3

Two long parallel wires

5 cm

apart carry

10 A

each in opposite directions. Find the force per unit length on each wire.

Solution:

$\frac{F}{L} = \frac{μ _{0} I _{1} I _{2}}{2 π d} = \frac{4 π \times 1 0 ^{- 7} \times 10 \times 10}{2 π \times 0.05} = \frac{2 \times 1 0 ^{- 4}}{0.05} = 4 \times 1 0^{- 4} N/m$

Direction: repulsive (currents antiparallel).

Solved Example 4

A wire PQ of mass

m

per unit length rests on two horizontal parallel rails carrying a constant current

I

. Below the rails, at distance

h

, a long straight wire carries current

I

in the opposite direction. Find the equilibrium separation

h

Figure 6Solved Example 4 (side view): wire

P Q

rests on the rails a height

h

above a long fixed wire. With antiparallel currents the force on

P Q

is repulsive (upward) and balances its weight:

\frac{μ _{0} I ^{2}}{2 π h} = m g

per unit length.

Solution:

Magnetic force per unit length on PQ (repulsive since the currents are antiparallel, so it acts upward) balances gravity:

$m g = \frac{μ _{0} I ^{2}}{2 π h} ⟹ h = \frac{μ _{0} I ^{2}}{2 π m g}$

This is a levitation setup - the wire "floats" above the current-carrying wire.

Solved Example 5

A rectangular loop of length

a

and width

b

carries current

I_{2}

and lies with its long sides parallel to a long straight wire (distance

d

from the nearer long side) carrying

I_{1}

in the same direction. Find the net force on the loop.

Figure 7Solved Example 5: rectangular loop (

a

along the wire,

b

across) beside a long wire. The near side carries

I_{2}

parallel to

I_{1}

and is attracted with

F_{1} = \frac{μ _{0} I _{1} I _{2} a}{2 π d}

; the far side is antiparallel and repelled with the smaller

F_{2} = \frac{μ _{0} I _{1} I _{2} a}{2 π ( d + b )}

. Forces on the two short sides cancel, so the net force points towards the wire.

Solution:

Near side (distance $d$ ) feels attractive force $F_{1} = \frac{μ _{0} I _{1} I _{2} a}{2 π d}$ (toward the wire).

Far side (distance $d + b$ ) feels repulsive force $F_{2} = \frac{μ _{0} I _{1} I _{2} a}{2 π ( d + b )}$ (away from the wire).

The two short ends (perpendicular to the wire) feel forces equal and opposite, so they cancel.

$F_{net} = F_{1} - F_{2} = \frac{μ _{0} I _{1} I _{2} a}{2 π} (\frac{1}{d} - \frac{1}{d + b}) = \frac{μ _{0} I _{1} I _{2} ab}{2 π d ( d + b )}$ , directed toward the wire.

6Point of Application

For calculating torque, the force on a straight current-carrying wire in a uniform field can be treated as acting at the midpoint of the wire. This is because the force is uniformly distributed along the length.

Common Mistakes to Avoid

Watch out

Integrating arc length for a curved wire in a uniform field, when the shortcut $L_{endpoints}$ works and gives the same answer instantly.
Forgetting a closed loop feels zero net force in a uniform field. If a question asks for the "force on a loop" in a uniform field, the answer is zero.
Missing the direction of parallel-wire force: same-direction currents attract, opposite-direction currents repel. Reverse of the intuition from charges.
Applying $F = B I L$ when $L$ is not perpendicular to $B$ : use $F = B I L sin θ$ or the vector form.
Ignoring the $sin θ$ factor when the wire and field are not perpendicular.
Confusing force on a wire with force between wires: " $F = B I L$ " is force on one wire in an external field; " $F / L = μ_{0} I_{1} I_{2} / (2 π d)$ " is the mutual force per unit length between two wires (with each wire in the other's field).
Not accounting for both long sides of a rectangular loop near a long wire: they feel opposite forces that partially cancel.

Frequently Asked Questions

Q1. What is the force on a current-carrying wire in a magnetic field?

For a straight wire of length $L$ carrying current $I$ in a uniform magnetic field $B$ , the force is $F = I (L \times B)$ , magnitude $F = B I L sin θ$ , where $θ$ is the angle between the current direction and $B$ . Direction: from the right-hand rule for $I \times B$ , or Fleming's left-hand rule.

Q2. Why is the net force on a closed loop zero in a uniform field?

For any wire in a uniform field, the net force equals $I (L_{endpoints} \times B)$ , where $L_{endpoints}$ is the straight-line vector from start to end of the wire. For a closed loop, start and end coincide, so $L_{endpoints} = 0$ and the net force vanishes. A loop can still feel a torque in a uniform field.

Q3. What is the force per unit length between two parallel current-carrying wires?

For two long parallel wires separated by $d$ , carrying $I_{1}$ and $I_{2}$ , the force per unit length is $F / L = μ_{0} I_{1} I_{2} / (2 π d)$ . Same-direction currents attract; opposite-direction currents repel.

Q4. Why do parallel currents attract each other?

Wire 1 creates a magnetic field circling around it. At the position of wire 2 (parallel to wire 1, same direction), this field points such that the force $I_{2} L_{2} \times B_{1}$ on wire 2 pulls it toward wire 1. Wire 1 experiences the same attraction toward wire 2 by symmetry (Newton's third law). Antiparallel currents give the reverse: repulsion.

Q5. How is the SI ampere defined?

Historically (pre-2019): one ampere is the current in each of two long, parallel wires $1$ m apart that produces a force of exactly $2 \times 1 0^{- 7}$ N per metre between them. After the 2019 SI redefinition, the ampere is defined via the exact elementary charge $e = 1.602176634 \times 1 0^{- 19}$ C; the parallel-wire relation still holds to high accuracy.

Q6. Does the shape of a wire between two endpoints affect the total force in a uniform field?

No. In a uniform field, only the straight vector from start to end of the wire matters: $F = I (L_{endpoints} \times B)$ . A straight wire, a zig-zag, and a curve with the same endpoints all feel the same total force. This is why any closed loop has zero net force.

Q7. How can a current-carrying wire be levitated by another current?

Place a horizontal wire directly above a parallel long straight wire, with the two currents in opposite directions. The upper wire feels an upward (repulsive) magnetic force per unit length $μ_{0} I_{1} I_{2} / (2 π d)$ . Setting this equal to its weight per unit length $m g$ gives the equilibrium height $d = μ_{0} I_{1} I_{2} / (2 π m g)$ .

Q8. What is Fleming's left-hand rule?

Stretch the thumb, index finger, and middle finger of your left hand at right angles. If the index finger points along the magnetic field $B$ , and the middle finger along the current $I$ , the thumb points in the direction of the force $F$ on the wire. It is a mnemonic equivalent to the right-hand rule for $I \times B$ .

Q9. Why does a rectangular current loop near a long straight wire feel a net force despite the "closed loop" rule?

The rule "net force on a closed loop is zero" applies only in a uniform field. The field near a long straight wire is non-uniform (varies as $1/ r$ ), so the two long sides of the rectangle sit in different field strengths. The near side feels a stronger force than the far side, leaving a net attractive (or repulsive) force toward (or away from) the straight wire.

Previous year questions on Magnetic Force on Current Carrying Wires

4 questions from past papers, each with a step-by-step solution.

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