Magnetic Force on a Current-Carrying Conductor

From a Single Charge to a Current Element

A current is just a stream of moving charges. Since each charge feels a force $q\vec{v}\times\vec{B}$ in a magnetic field, a wire carrying current must feel a net force too — the sum of forces on all its charge carriers.

Let's derive it cleanly. Consider a short straight piece of wire of length $dl$ and cross-section $A$ carrying a steady current $I$. Let $n$ be the number density of charge carriers (electrons in a metal, say), each with charge $q$ and drift velocity $\vec{v}_d$.

The total number of carriers in this small element is $N = n\,A\,dl$. The force on each carrier is $q\vec{v}_d\times\vec{B}$, so the net force on the element is

$$d\vec{F} = N\,(q\,\vec{v}_d\times\vec{B}) = (nA\,dl)\,(q\,\vec{v}_d\times\vec{B})$$

Now recall from Chapter 3 that the current is $I = n\,q\,v_d\,A$. Defining $d\vec{l}$ as a tiny length vector pointing along the direction of conventional current flow (same as $\vec{v}_d$ for positive carriers), we get

$$\boxed{\;d\vec{F} = I\,d\vec{l}\times\vec{B}\;}$$

This is the force on a current element — the workhorse equation of this section. Notice how the messy details (number of carriers, their drift speed) all collapsed into the macroscopic quantity $I$.

Note: Here $\vec{v}_d$ refers to the drift velocity of the positive charge carriers so that $d\vec{l}$ aligns with the conventional current. In metals where electrons (negative charges) move opposite to the current, $d\vec{l}$ still points along the current direction.

Force on a Straight Wire in a Uniform Field

For a straight wire of length $L$ carrying current $I$ in a uniform magnetic field $\vec{B}$, we can integrate $d\vec{F} = I\,d\vec{l}\times\vec{B}$ trivially:

$$\boxed{\;\vec{F} = I\,\vec{L}\times\vec{B}\;}$$

where $\vec{L}$ is a vector of length $L$ pointing along the direction of current flow.

Magnitude

$$F = B\,I\,L\,\sin\theta$$

where $\theta$ is the angle between $\vec{L}$ and $\vec{B}$.

• If the wire is parallel to $\vec{B}$ ($\theta = 0°$): $F = 0$.
• If the wire is perpendicular to $\vec{B}$ ($\theta = 90°$): $F = BIL$ — maximum.

Direction

Perpendicular to both the wire and $\vec{B}$, given by the right-hand rule for $\vec{L}\times\vec{B}$ (or equivalently Fleming's Left-Hand Rule, below).

[JEE Tip] When the wire is not straight (curved or bent), you cannot use $\vec{F} = I\vec{L}\times\vec{B}$ directly — you must integrate $d\vec{F} = I\,d\vec{l}\times\vec{B}$ along the wire. For a uniform field, however, this integration has a beautiful shortcut (next note).

Fleming's Left-Hand Rule

A handy mnemonic for the direction of force on a current-carrying wire: Fleming's Left-Hand Rule.

Stretch the forefinger, middle finger, and thumb of your left hand so they are mutually perpendicular:

Finger	Represents
Forefinger	Direction of magnetic field $\vec{B}$
Middle finger	Direction of conventional current $I$
Thumb	Direction of force $\vec{F}$ on the wire

Mnemonic: FBI Rule — Forefinger for B (Field), Index for I (Current), and Thumb for F (Force).

Force on a Closed Loop in a Uniform Field — Why It's Zero

Here's a beautiful result that students often miss: the net force on a closed current loop in a uniform magnetic field is zero.

Proof (one line)

$$\vec{F}_{net} = \oint I\,d\vec{l}\times\vec{B} = I\Bigl(\oint d\vec{l}\Bigr)\times\vec{B} = I\,(\vec{0})\times\vec{B} = \vec{0}$$

because the closed-loop integral of $d\vec{l}$ vanishes — the displacement vector returns to where it started.

What this does not mean

The loop can still experience a net torque even though the net force is zero. That's how electric motors work — a uniform $\vec{B}$ produces a torque on a current loop, spinning it. We'll handle that fully in Section 10.

Where it matters

This zero-force result illustrates how summing all segment forces can give zero net translation; it is distinct from the force between parallel currents, which we will derive formally in Section 9.

[JEE Tip] Two consequences worth remembering:

1. In a uniform field, only torque acts on a loop, not net force.
2. In a non-uniform field, a current loop does experience a net force — this is how an electromagnet attracts iron, and how magnetic confinement works in plasmas.

Memory Capsule

Lock in these results before moving to motion in a field (Section 4).

The two master equations

Object	Force
Current element $d\vec{l}$	$d\vec{F} = I\,d\vec{l}\times\vec{B}$
Straight wire of length $\vec{L}$ in uniform $\vec{B}$	$\vec{F} = I\,\vec{L}\times\vec{B}$

Magnitude

$$F = BIL\sin\theta$$

where $\theta$ is the angle between the wire and the field.

Direction — Fleming's Left-Hand Rule

Finger	Quantity
Forefinger	$\vec{B}$ (Field)
Middle	$I$ (Current)
Thumb	$\vec{F}$ (Force)

Special cases

• Wire parallel to $\vec{B}$: $F = 0$.
• Wire perpendicular to $\vec{B}$: $F = BIL$ (maximum).
• Wire is curved: integrate $d\vec{F}$ along the wire. In a uniform field, the result for a wire from point A to point B depends only on the straight vector $\vec{L}_{AB}$, not on the path.

Closed loop in a uniform field

• Net force = 0 (always, for any shape).
• Net torque may be non-zero — that's the motor principle (Section 10).
• In a non-uniform field, a closed loop does experience a net force (electromagnet pulling on iron).

One-line takeaway

The magnetic field pushes wires the same way it pushes charges — but through the macroscopic handle of current, $I\vec{L}\times\vec{B}$.

Solved Examples

Example 1: Force on a horizontal current-carrying wire

A horizontal wire of length $L = 0.5$ m carries a current $I = 4$ A from east to west. It lies in a region where $\vec{B} = 0.3$ T points vertically downward. Find the magnitude and direction of the magnetic force on the wire.

Solution. Use $\vec{F} = I\vec{L}\times\vec{B}$.

Magnitude: $$F = BIL\sin 90° = (0.3)(4)(0.5) = 0.6\;\text{N}.$$
Direction: Fleming's Left-Hand Rule — forefinger down (along $\vec{B}$), middle finger west (along $I$), thumb points south.

Example 2: Wire balanced against gravity

A horizontal wire of mass $m = 50$ g and length $L = 25$ cm carries a current $I$ in a uniform horizontal magnetic field $B = 0.2$ T (perpendicular to the wire). What current is required to make the magnetic force on the wire exactly support its weight? Take $g = 10$ m/s².

Solution. For levitation, magnetic force = weight:

$$BIL = mg \quad\Rightarrow\quad I = \frac{mg}{BL}$$

$$I = \frac{(0.05)(10)}{(0.2)(0.25)} = \frac{0.5}{0.05} = 10\;\text{A}.$$
(The direction of current must be chosen so that $I\vec{L}\times\vec{B}$ points upward — left-hand rule decides which way.)

Example 3: Wire at an angle

A straight wire of length $0.4$ m carries $5$ A in a uniform magnetic field of $0.6$ T. The wire makes an angle of $37°$ with $\vec{B}$. Find the force on the wire. ($\sin 37° = 0.6$.)

Solution. Use $F = BIL\sin\theta$.

$$F = (0.6)(5)(0.4)(0.6) = 0.72\;\text{N}.$$
The direction is perpendicular to the plane containing the wire and $\vec{B}$.

Example 4: Force on a curved wire in a uniform field

A semicircular wire of radius $R$ carrying current $I$ lies in the plane of the page. A uniform magnetic field $\vec{B}$ points perpendicular to the plane (into the page). The wire's two ends are joined to the rest of a circuit by straight leads. Find the net force on the semicircular portion alone.

Solution. Here's a beautiful shortcut. In a uniform field,

$$\vec{F}_{semi} = I\Bigl(\int d\vec{l}\Bigr)\times\vec{B} = I\,\vec{L}_{AB}\times\vec{B}$$

where $\vec{L}_{AB}$ is the straight chord from the start of the semicircle to its end. For a semicircle of radius $R$, that chord has length $2R$.

$$F = I(2R)B = 2IRB$$

Direction: perpendicular to the chord $AB$, in the plane of the page, given by the right-hand rule on $\vec{L}_{AB}\times\vec{B}$.
Note: this trick works only because the field is uniform. The result is independent of how curved the wire is — only the endpoints matter.

Example 5: Closed loop in a uniform field

A square loop of side $0.2$ m carries a current of $3$ A in a uniform field $\vec{B} = 0.5$ T perpendicular to the loop's plane. Find the net force on the loop.

Solution. The net force on any closed loop in a uniform field is zero:

$$\vec{F}_{net} = I\Bigl(\oint d\vec{l}\Bigr)\times\vec{B} = \vec{0}$$

So $F_{net} = 0$ regardless of the current, side length, or field magnitude. (The loop may still experience a non-zero torque — that's a different question, treated in Section 10.)

Example 6: Power line in Earth's field

A horizontal transmission line carries $I = 100$ A from north to south. The horizontal component of Earth's magnetic field at that location is $B_H = 4\times 10^{-5}$ T, pointing north. Find the force per unit length on the wire.

Solution. Force per unit length: $f = BI\sin\theta$.

Here current direction is south, field direction is north (anti-parallel, $\theta = 180°$), so

$$f = BI\sin 180° = 0$$

No magnetic force from the horizontal component. Only the vertical component of Earth's field would produce a force on this wire — which is small ($B_V \sim 4\times 10^{-5}$ T at many locations, giving $f \sim 4\times 10^{-3}$ N/m horizontally east or west).

Example 7: Current-carrying wire in a field at $30°$

A wire of length $1.5$ m carries a current of $2.5$ A. It is placed in a uniform magnetic field of $0.4$ T such that the wire makes an angle of $30°$ with the field. Calculate the force on the wire.

Solution. Use $F = BIL\sin\theta$.

$$F = (0.4)(2.5)(1.5)\sin 30° = (0.4)(2.5)(1.5)(0.5) = 0.75\;\text{N}$$

Example 8: Maximum and minimum force

A wire of length $0.6$ m carrying $5$ A is placed in a magnetic field of $0.8$ T. Find (a) the maximum possible force on the wire (b) the minimum possible force.

Solution. Use $F = BIL\sin\theta$.

(a) Maximum at $\theta = 90°$ (wire perpendicular to $\vec{B}$):

$$F_{max} = BIL = (0.8)(5)(0.6) = 2.4\;\text{N}$$

(b) Minimum at $\theta = 0°$ or $180°$ (wire parallel or anti-parallel to $\vec{B}$):

$$F_{min} = 0$$

Takeaway: by rotating the orientation of the wire, you can tune the force from $0$ to $BIL$ continuously.