MHT-CET Physics · Magnetic Fields Due to Electric Current

Force on a Current-Carrying Wire, and Between Parallel Wires

A wire of length L carrying current I in a field B feels F = I(L × B); two long parallel wires each sit in the other's field, so they attract when their currents run the same way and repel when opposite, with a force per unit length μ₀I₁I₂/(2πd).

All Magnetic Fields Due to Electric Current notes MHT-CET Physics strategy

Why this matters

14 PYQs, 6 HARD. Three are the force on a single conductor — a wire in a field given as a vector, one side of a triangular loop, and the net force on a square coil beside a long wire; eleven are parallel wires — how the force changes with currents and distance, attraction or repulsion, three wires side by side, a wire floating above another, and the currents from the fields at the midpoint. Two cards.

Concept 1 of 2: Force on a Straight Conductor

F = I(L × B), of size ILB sin θ where θ is the angle between the wire and the field. For a field given as a vector, work the cross product; only the parts of B across the wire count. A closed loop in a UNIFORM field feels no net force, but near a long straight wire the field is not uniform: the side nearer the wire feels a stronger push than the far side, and the sides perpendicular to the wire cancel.

Definition

$\vec F = I\vec L \times \vec B$ , $|F| = ILB\sin\theta$ .
Wire along x, $\vec B = B_0(\hat i - \hat j - \hat k)$ : $|F| = \sqrt{2}ILB_0$ .
5 cm side of a 5-12-13 triangle with B along the 13 cm side: $\sin\theta = \tfrac{12}{13}$ , $F = \tfrac{9}{130}$ N at 2 A, 0.75 T.
Square coil (side L) beside a long wire, near side at L/3: $\dfrac{\mu_0I_1I_2}{2\pi}\left(\dfrac{L}{L/3} - \dfrac{L}{4L/3}\right) = \dfrac{9\mu_0I_1I_2}{8\pi}$ .

Force on a wire

\vec F = I\,\vec L \times \vec B

Worked example

A 0.5 m wire carries 4 A at 30° to a 0.2 T field. Force on it?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q14Hard

Example 1 · Magnetic Fields Due to Electric Current · Force on Current-Carrying Conductor and Parallel Wires

A square coil ABCD of side '

L

' is carrying a current in clockwise direction. A straight conductor carrying current

I_{2}

(upward direction) is kept parallel to side AB at a distance

\frac{L}{3}

in the plane of ABCD. The net force on the coil ABCD is (

\mu_{0}=

magnetic permeability)

Expecting the near and far sides to cancel

They carry opposite currents but sit at different distances from the wire, where the field differs. Only the perpendicular sides cancel; the net force is the difference of the two parallel sides.

Drill 2 more on force on a straight conductor

Concept 2 of 2: Parallel Wires: Attraction, Repulsion and Balance

Each wire sits in the other's field μ₀I/(2πd), so the force per unit length is μ₀I₁I₂/(2πd): like currents attract, opposite currents repel. It scales with each current and inversely with distance — doubling both currents quadruples it, halving the distance as well makes it eight times. A wire between two others feels no force where the two pulls balance: I₁/x = I₂/(d − x). A light wire above a heavy one with opposite currents floats where the repulsion μ₀I²l/(2πh) equals its weight.

Definition

$\dfrac{F}{l} = \dfrac{\mu_0I_1I_2}{2\pi d} = \dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2}{d}$ ; same direction ⇒ attract, opposite ⇒ repel.
Currents × 2 ⇒ F × 4; also d ÷ 2 ⇒ F × 8. One current × 2 and reversed, d × 3 ⇒ $-\tfrac{2}{3}F$ .
Heater leads: $I = P/V$ (1 kW, 100 V ⇒ 10 A; 2 mm apart ⇒ $10^{-2}$ N/m).
Floating wire: $\dfrac{\mu_0I^2l}{2\pi h} = mg$ (25 A, 1 m, 2.5 g ⇒ 5 mm).
No-force position between wires D (15 A) and B (10 A), 15 cm apart: $\tfrac{15}{x} = \tfrac{10}{15 - x}$ ⇒ x = 9 cm.
From midpoint fields: same direction gives $B_1 - B_2$ , one reversed gives $B_1 + B_2$ ; solve for the ratio of currents.

Force between parallel wires

\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}

Worked example

Wires 4 cm apart carry 6 A and 8 A in the same direction. Force per metre, and its sense?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q34Hard

Example 2 · Magnetic Fields Due to Electric Current · Force on Current-Carrying Conductor and Parallel Wires

Three long, straight parallel wires carrying currents are arranged as shown. The wire C which carries a current of 5.0 A is so placed that it experiences no force. The distance of wire C from wire D is

Getting attraction and repulsion backwards

Parallel currents in the SAME direction attract; opposite currents repel. It is the reverse of like charges.

Drill 10 more on parallel wires: attraction, repulsion and balance

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Force on a Straight Conductor
Force on a wire
$\vec F = I\,\vec L \times \vec B$
Parallel Wires: Attraction, Repulsion and Balance
Force between parallel wires
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$

Watch out for (2)

Expecting the near and far sides to cancel
→ Force on a Straight Conductor
Getting attraction and repulsion backwards
→ Parallel Wires: Attraction, Repulsion and Balance

Test yourself on Magnetic Fields Due to Electric Current

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Related notes

Field of a Current — the field each wire makes →