Magnetic force

Bend a beam of electrons with nothing but a horseshoe magnet and you have already performed one of the cleanest experiments in physics: a magnetic field visibly steers a moving charge without ever touching it, without doing a single joule of work, and without changing its speed by even a hair. That single, strange fact — a force that turns but never accelerates along the direction of travel — is the seed from which this entire chapter grows. A television tube once used it to paint pictures line by line; a mass spectrometer uses it today to sort atoms by mass; the aurora borealis is the sky itself demonstrating it on a planetary scale as charged particles from the Sun spiral down along Earth's field lines. Everything in Moving Charges and Magnetism after Oersted's discovery that currents make magnetic fields comes down to answering the reverse question: what does a magnetic field do back to a moving charge, and to the current-carrying wires built from countless such charges? This article is the hinge of the chapter. It starts from the single-charge law $\vec F = q\vec v \times \vec B$ , already met in the previous article, and builds outward to the force on a straight current-carrying conductor, $\vec F = I\vec l \times \vec B$ , and to the force between two parallel current-carrying wires — the very relation that defines the ampere, the SI base unit of current. By the end you should be able to move fluently between the microscopic picture (one charge, one velocity) and the macroscopic picture (a wire full of charges, one current), and calculate correctly in both. We keep the mathematics to vectors, cross products, and right-angle trigonometry — no calculus is needed anywhere in this article — while being exact about magnitudes, directions, and the handful of numbers CBSE expects you to compute quickly and correctly under exam pressure.

Key takeaways

A magnetic field exerts a force on a charge only while it is moving, and only on the part of its velocity perpendicular to $\vec B$ : $\vec F = q\vec v \times \vec B$ , magnitude $F = qvB\sin\theta$ .
This force is always perpendicular to $\vec v$ , so it does zero work — it can bend a path into a circle but can never change a charge's speed or kinetic energy.
A current-carrying wire feels a force too, because it is just billions of moving charges acting together: $\vec F = I\vec l \times \vec B$ , magnitude $F = BIL\sin\theta$ .
Two parallel wires carrying current exert forces on each other — same direction currents attract, opposite direction currents repel — with force per unit length $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ .
This parallel-wire formula is not just a curiosity — it is the operational definition of the ampere, the SI base unit of electric current.
Direction in every case comes from the right-hand rule applied to a cross product, with an extra sign-flip for negative charges.
Key idea: a magnetic field pushes sideways on any moving charge (F = qvB sinθ) and, by the same logic applied to a wire full of charges, sideways on any current-carrying conductor (F = BIL sinθ), with the force between two such conductors defining the ampere itself.

The core insight

The magnetic force on a moving charge, $\vec F = q\vec v \times \vec B$ , is always perpendicular to velocity and does no work; summing this force over every charge carrier in a wire gives the force on a current, $\vec F = I\vec l \times \vec B$ , whose parallel-wire version, $F/L = \mu_0 I_1 I_2/2\pi d$ , is used to define the ampere.

Where it fits in the syllabus

This article sits at the start of CBSE Class 12 Physics unit Moving Charges and Magnetism, immediately after Oersted's qualitative demonstration that currents produce magnetic fields. Having established that a field exists, the syllabus turns the question around and asks what that field does to charges and currents — the force topics covered here (force on a moving charge, force on a current-carrying conductor, and force between two parallel conductors) are explicitly listed in the prescribed scope, along with the resulting definition of the ampere. What is deliberately kept out at this stage is anything about the source of the field beyond Oersted's qualitative picture — Biot–Savart law and Ampere's circuital law are separate articles later in the same chapter — and any full quantitative treatment of circular motion (cyclotron-style) is likewise outside this chapter's core numericals. Everything here uses only vectors, the cross product, and right-angle trigonometry, matching the board's algebra-and-vectors approach with no calculus.

The physical picture

A magnetic field $\vec B$ does not push on a charge sitting still, and it does not push along the direction the charge is already moving — it only pushes on the part of the charge's motion that is across the field lines. For a charge $q$ moving with velocity $\vec v$ through a field $\vec B$ , the force is the vector cross product

\vec F = q\,(\vec v \times \vec B)

where: $q$ is the charge (with its sign — positive or negative), $\vec v$ is the instantaneous velocity, and $\vec B$ is the magnetic field at that point; the result $\vec F$ automatically comes out perpendicular to both $\vec v$ and $\vec B$ , because that is what a cross product always produces. In magnitude, if $\theta$ is the angle between $\vec v$ and $\vec B$ ,

F = qvB\sin\theta

Read this physically before you ever plug in numbers: the $\sin\theta$ factor is the whole story. When $\vec v$ is parallel to $\vec B$ ( $\theta = 0°$ ), the charge glides along the field lines untouched, $F = 0$ . When $\vec v$ is perpendicular to $\vec B$ ( $\theta = 90°$ ), the force is at its strongest, $F = qvB$ . Every other angle interpolates smoothly in between. The direction itself follows the right-hand rule: point the fingers along $\vec v$ , curl them toward $\vec B$ , and the thumb gives $\vec v \times \vec B$ for a positive charge — for a negative charge, reverse that direction.

Figure

Two diagrams side by side showing a positive charge and a negative charge moving to the right through a magnetic field pointing into the page, with the resulting magnetic force arrow pointing upward for the positive charge and downward for the negative charge

The same logic scales up naturally. A current-carrying wire is nothing more than an enormous number of moving charges confined to a conductor, so it should feel a magnetic force too — and it does. For a straight segment of wire of length $l$ carrying current $I$ in a field $\vec B$ ,

\vec F = I\,\vec l \times \vec B, \qquad F = BIl\sin\theta

where: $I$ is the current, $\vec l$ is a vector along the wire in the direction of conventional current flow with magnitude equal to the wire's length, $\vec B$ is the field, and $\theta$ is the angle between the wire and the field. The direction, once again, comes from the right-hand rule applied to $\vec l \times \vec B$ — no separate sign flip is needed here because conventional current direction already accounts for the sign of the actual charge carriers. Two such current-carrying wires running parallel to each other each sit in the other's magnetic field and therefore exert forces on each other; per unit length,

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

where: $I_1$ and $I_2$ are the two currents, $d$ is the perpendicular separation between the wires, and $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space. Currents flowing in the same direction attract; currents flowing in opposite directions repel — the reverse of what many students expect from an analogy with like electric charges repelling. This exact relation is how the ampere itself is operationally defined: one ampere is the current that, flowing in each of two infinitely long parallel wires one metre apart in vacuum, produces a force of exactly $2\times10^{-7}\ \text{N}$ on each metre of either wire.

Derivation ABuilding the force on a wire from the force on one charge

The goal is to show that the macroscopic force law on a current-carrying wire, $F = BIl\sin\theta$ , is nothing but the microscopic single-charge law $F = qvB\sin\theta$ added up over every charge carrier drifting through the wire — so that a student who trusts $\vec F = q\vec v\times\vec B$ can reconstruct the conductor formula from scratch in an exam without memorising it as a separate, disconnected fact, and can see clearly why the direction rule for a wire is really the same right-hand rule used for a single charge.

Step 1. Picture a straight wire of length $l$ and cross-sectional area $A$ , carrying current $I$ , sitting in a uniform field $\vec B$ at angle $\theta$ to the wire. Inside the wire, free charge carriers (electrons in a metal, but we can reason with an effective positive carrier of charge $q$ per carrier, since current direction already tracks the correct sign) drift along the wire with an average drift speed $v_d$ . Each individual carrier, simply by virtue of moving with speed $v_d$ through $\vec B$ , feels the single-charge force $F_{\text{one}} = q v_d B \sin\theta$ , directed perpendicular to the wire — this is just the law from the previous section, applied to one carrier.

Step 2. Count how many carriers are inside the wire at once. If $n$ is the number of free charge carriers per unit volume (a property of the metal), the volume of the wire segment is $A l$ , so the total number of carriers in that length is $N = nAl$ .

Step 3. Add up the force on every carrier. Since every carrier drifts with the same $v_d$ and feels the same $F_{\text{one}}$ in the same direction, the total force on the segment is simply $N$ times the single-carrier force: $F = N q v_d B \sin\theta = (nAl)\,q v_d B \sin\theta$ .

Step 4. Recognise the current hiding inside that expression. The defining relation for drift current is $I = nqv_dA$ — this is exactly the combination $nqv_dA$ that appears (rearranged) inside our result. Substituting, $F = (nqv_dA)\,l\,B\sin\theta = I\,l\,B\sin\theta$ , which is precisely $F = BIl\sin\theta$ , the force law on a current-carrying conductor. The direction of $\vec F$ is still given by the same right-hand rule as before, now applied between the current direction $\vec l$ and $\vec B$ , because the drift velocity direction and the conventional current direction coincide by definition. Verified: substituting $I = nqv_dA$ into $F = nAl\,qv_dB\sin\theta$ collapses cleanly to $F = BIl\sin\theta$ with no leftover microscopic quantities ( $n$ , $q$ , $v_d$ , $A$ ) — confirming the macroscopic law is exactly the sum of the microscopic one.

Worked examples

Board level - force on a current-carrying wire

A straight wire of length $L = 2\ \text{m}$ carries a current $I = 5\ \text{A}$ , placed perpendicular to a uniform magnetic field of magnitude $B = 0.3\ \text{T}$ . Find the magnetic force on the wire.

Step 1. Identify the angle. The wire is perpendicular to the field, so $\theta = 90°$ and $\sin\theta = 1$ — the same maximum-force condition used for the perpendicular proton of Worked example 1 in the previous article, now applied to a wire instead of a single charge.

Step 2. Apply the formula. $F = BIL\sin\theta = (0.3)(5)(2)(1)$ .

Step 3. Compute. $F = 0.3 \times 5 \times 2 = 3\ \text{N}$ .

Step 4. State the direction. Using the right-hand rule on $\vec l \times \vec B$ with $\vec l$ along the current direction, the force on the wire points perpendicular to both the current and the field — in an exam, sketch the wire, field, and force as three mutually perpendicular directions to earn the direction mark.

The force on the wire is $F = 3\ \text{N}$ , directed perpendicular to both the current and the field.

Competitive level - force between two current-carrying wires

Two long, straight, parallel wires are separated by $d = 5\ \text{cm} = 0.05\ \text{m}$ and carry currents $I_1 = 6\ \text{A}$ and $I_2 = 4\ \text{A}$ in the same direction. Find the force per unit length between them, state whether it is attractive or repulsive, and find the total force on a $2\ \text{m}$ length of either wire.

Step 1. Write the formula and identify knowns. $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ , with $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ , $I_1 = 6\ \text{A}$ , $I_2 = 4\ \text{A}$ , $d = 0.05\ \text{m}$ .

Step 2. Simplify the constant first — this is the step competitive exams reward for speed. Since $\dfrac{\mu_0}{2\pi} = \dfrac{4\pi\times10^{-7}}{2\pi} = 2\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ , the formula collapses to $\dfrac{F}{L} = \dfrac{(2\times10^{-7})I_1I_2}{d}$ , avoiding repeated $\pi$ arithmetic.

Step 3. Substitute and compute the force per unit length. $\dfrac{F}{L} = \dfrac{(2\times10^{-7})(6)(4)}{0.05} = \dfrac{4.8\times10^{-6}}{0.05} = 9.6\times10^{-5}\ \text{N/m}$ .

Step 4. Determine the nature of the force and scale to the given length. Because both currents flow in the same direction, the wires attract each other. For $L = 2\ \text{m}$ : $F = (9.6\times10^{-5}\ \text{N/m})(2\ \text{m}) = 1.92\times10^{-4}\ \text{N}$ .

The force per unit length is $9.6\times10^{-5}\ \text{N/m}$ (attractive), giving $F = 1.92\times10^{-4}\ \text{N}$ over a 2 m length. Timing note: pre-simplify $\mu_0/2\pi$ to $2\times10^{-7}$ before touching the given numbers — this single habit turns a two-minute calculation into a fifteen-second one and is exactly the kind of arithmetic shortcut competitive exams are timed to reward.

Common traps

Students who understand the physics perfectly still lose marks on this topic for a handful of repeatable reasons. Forgetting the $\sin\theta$ factor is the single biggest one — writing $F = qvB$ or $F = BIL$ unconditionally, even when the angle given is not $90°$ , throws away marks on a question that was specifically testing whether you noticed the angle. A close cousin is silently assuming $\theta$ is always the angle "between the wire and the diagram's horizontal line" rather than the actual angle between the current direction and $\vec B$ — always re-read which two vectors the angle is between. Another frequent slip is flipping the direction rule: the right-hand rule for $\vec v\times\vec B$ or $\vec l\times\vec B$ gives one direction, but for a negative charge the force must be reversed, and students often apply the rule correctly and then forget the flip. Finally, in the parallel-wire problem, many students reflexively apply the "like charges repel" intuition from electrostatics and conclude that same-direction currents should repel — magnetism does not obey that analogy; same-direction currents attract.

You should be able to…

State $\vec F = q\vec v\times\vec B$ and $F = qvB\sin\theta$ , and correctly identify when the force is zero, maximum, or in between.
Extend the single-charge law to a current-carrying wire, $\vec F = I\vec l\times\vec B$ and $F = BIl\sin\theta$ , and explain in one sentence why the two formulas are really the same physics.
State the parallel-wire force law $F/L = \mu_0 I_1I_2/2\pi d$ , correctly judge attraction versus repulsion from the current directions, and connect it to the definition of the ampere.
Apply the right-hand rule confidently to find force direction for a positive charge, a negative charge, and a current-carrying wire, without mixing up the sign flip.
Explain precisely why the magnetic force does zero work and can never change a moving charge's speed or kinetic energy.
Simplify $\mu_0/2\pi = 2\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ from memory to speed up parallel-wire numericals.

Ask the tutor: Still shaky on why the same right-hand rule handles a single moving charge, a whole current-carrying wire, and two wires pushing on each other — ask the AI tutor to walk through all three with one consistent hand-and-vector picture.

Why does the force formula use v cross B instead of just multiplying v and B together?

Because the magnetic force depends on direction, not just size — a charge moving straight along the field lines feels nothing, while one moving straight across them feels the maximum push, and every angle in between gives a value scaled by sine of that angle. A plain multiplication like $qvB$ has no way to encode that angle-dependence or the resulting perpendicular direction. The cross product $\vec v\times\vec B$ automatically produces both the correct $\sin\theta$ magnitude and a direction perpendicular to both vectors in one compact piece of notation, which is exactly the geometry the real force obeys.

If the magnetic force does no work, how does a magnet ever lift or move anything, like a fridge magnet holding paper?

That everyday pull comes from a different mechanism, not the $\vec F = q\vec v\times\vec B$ law for a single free charge. A magnet attracting another magnet or a piece of iron involves forces between magnetic dipoles and induced magnetisation inside the material, which is a separate topic (covered under "Magnetism and Matter") built from many microscopic current loops, not from a single charge drifting through empty space. Within this chapter, whenever you see $\vec F = q\vec v\times\vec B$ or $\vec F = I\vec l\times\vec B$ , it always refers to a charge or current already in motion inside a field created elsewhere.

Why do current-carrying wires attract when currents run the same way, when like electric charges repel?

It feels backwards, but the underlying reason is genuinely different physics. Each wire creates its own magnetic field circling around it (a separate topic, Biot–Savart law and Ampere's law), and the second wire's current sits inside that field and feels a force from it via $\vec F = I\vec l\times\vec B$ . Working through the geometry of the right-hand rule for two same-direction currents shows the resulting force vectors point toward each other, not away — it is not governed by any "like repels like" rule from electrostatics, so it is best to simply memorise the result (same direction attracts, opposite direction repels) rather than force an analogy that does not hold.

Is the circular path of a charged particle in a magnetic field something I need to calculate for this chapter?

The 2025-26 CBSE syllabus for this chapter keeps the treatment of a charged particle's motion in a field mostly qualitative at the "magnetic force" level — you should know that a charge moving perpendicular to a uniform field follows a circular arc because the force is always perpendicular to velocity and constant in magnitude, which is exactly the condition for circular motion. A full quantitative cyclotron-style treatment with orbit radius formulas is not the core numerical focus of this article; if your textbook or teacher goes further, treat it as enrichment on top of the force law you have learned here, not a replacement for it.