Sources and Fields
The starting point is the parallel between electricity and magnetism set out in NCERT Section 4.2.1. A static charge $Q$ is the source of an electric field $\mathbf{E}$, and a second charge $q$ placed in that field experiences a force $\mathbf{F} = q\mathbf{E}$. The field is not merely a mathematical artefact; it carries energy and momentum and obeys the principle of superposition, so the field of several charges adds vectorially.
Magnetism extends this picture. Just as static charges produce an electric field, currents or moving charges produce, in addition, a magnetic field $\mathbf{B}(\mathbf{r})$ — again a vector field defined at every point in space, and again obeying superposition. The decisive difference appears when we ask how a charge responds to $\mathbf{B}$: the response depends not only on the charge but on how fast and in what direction it is moving.
The Lorentz Force Law
Consider a point charge $q$ moving with velocity $\mathbf{v}$ at a position $\mathbf{r}$ in the presence of both an electric field $\mathbf{E}(\mathbf{r})$ and a magnetic field $\mathbf{B}(\mathbf{r})$. The total force on it is
$$\mathbf{F} = q\left[\mathbf{E}(\mathbf{r}) + \mathbf{v}\times\mathbf{B}(\mathbf{r})\right] \equiv \mathbf{F}_{\text{electric}} + \mathbf{F}_{\text{magnetic}}$$
This expression was given first by H.A. Lorentz, building on the extensive experiments of Ampère and others, and is called the Lorentz force. The first term $q\mathbf{E}$ is the familiar electric force. The second term $q(\mathbf{v}\times\mathbf{B})$ is the magnetic force, and NCERT highlights three features that define its character.
| Feature of the magnetic force | Consequence |
|---|---|
| Depends on $q$, $\mathbf{v}$ and $\mathbf{B}$ | Force on a negative charge is opposite to that on a positive charge in the same field and velocity. |
| Contains the vector product $\mathbf{v}\times\mathbf{B}$ | The force vanishes when $\mathbf{v}$ is parallel or anti-parallel to $\mathbf{B}$; it acts sideways, perpendicular to both. |
| Requires motion of the charge | The magnetic force is zero if the charge is stationary, since then $|\mathbf{v}| = 0$. Only a moving charge feels it. |
Magnitude and Direction
Writing the magnetic force in component form, $\mathbf{F} = q\,\mathbf{v}\times\mathbf{B} = qvB\sin\theta\,\hat{\mathbf{n}}$, where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{B}$ and $\hat{\mathbf{n}}$ is the unit vector perpendicular to the plane containing both. The magnitude is therefore
$$F = qvB\sin\theta$$
The $\sin\theta$ factor encodes the most-tested behaviour of the force: it is greatest when the charge moves at right angles to the field ($\theta = 90^\circ$) and exactly zero when it moves along the field ($\theta = 0^\circ$ or $180^\circ$). The direction of $\mathbf{v}\times\mathbf{B}$ is fixed by the screw rule, or equivalently the right-hand rule for a vector product, illustrated below.
Zero force when v is parallel to B
A charge fired along the field lines feels no magnetic force at all, because $\sin 0^\circ = 0$. Students who memorise "$F = qvB$" without the $\sin\theta$ wrongly compute a non-zero force for a charge moving parallel to $\mathbf{B}$.
Always write $F = qvB\sin\theta$. Maximum force at $\theta = 90^\circ$; zero force at $\theta = 0^\circ$ or $180^\circ$.
A proton enters a 0.5 T field at $30^\circ$ to the field lines with speed $4\times10^{6}\,\text{m/s}$. Find the magnetic force on it. ($q = 1.6\times10^{-19}\,\text{C}$)
$F = qvB\sin\theta = (1.6\times10^{-19})(4\times10^{6})(0.5)(\sin 30^\circ)$.
$F = (1.6\times10^{-19})(4\times10^{6})(0.5)(0.5) = 1.6\times10^{-13}\,\text{N}$, directed perpendicular to both $\mathbf{v}$ and $\mathbf{B}$.
Why Magnetic Force Does No Work
Because $\mathbf{F}_{\text{magnetic}} = q(\mathbf{v}\times\mathbf{B})$ is always perpendicular to $\mathbf{v}$, it can do no work on the charge. Work is the dot product of force and displacement, and the displacement at each instant lies along $\mathbf{v}$; since the force has no component in that direction, the work is zero. The magnetic force can therefore change only the direction of the velocity, never its magnitude.
The immediate physical consequence is that the speed and kinetic energy of a charged particle remain constant in a purely magnetic field. This is exactly why a charge entering a uniform field at right angles travels in a circle of constant radius rather than spiralling inward or speeding up — the basis of circular motion and the cyclotron.
Magnetic force changes direction, not speed
A common error is to assume a magnetic field can accelerate a particle and increase its kinetic energy. It cannot. Only an electric field (the $q\mathbf{E}$ term) can do work and change the speed; the magnetic term merely deflects.
In a magnetic field alone: $|\mathbf{v}|$ = constant, KE = constant, only the direction of motion changes.
Because the magnetic force is perpendicular and does no work, a charge perpendicular to $\mathbf{B}$ moves in a circle. See Motion in a Magnetic Field and the Cyclotron for the radius, period and accelerator design.
The Tesla and the Gauss
The force expression also defines the unit of magnetic field. Setting $q$, $F$ and $v$ all equal to unity in $F = qvB\sin\theta$ with the charge moving perpendicular to $\mathbf{B}$, the magnitude of $\mathbf{B}$ is 1 SI unit when the force on a unit charge (1 C) moving perpendicular to $\mathbf{B}$ at 1 m/s is one newton. Dimensionally $[B] = [F/qv]$, giving units of newton-second per coulomb-metre.
| Unit | Definition / value | Notes |
|---|---|---|
| Tesla (T) | 1 T = 1 N·s/(C·m) = 1 N/(A·m) |
SI unit, named after Nikola Tesla (1856–1943). A rather large unit. |
| Gauss (G) | 1 G = 10⁻⁴ T |
Smaller non-SI unit, often used in practice. |
| Earth's field | ≈ 3.6 × 10⁻⁵ T |
Order-of-magnitude reference for comparison. |
Force on a Current-Carrying Conductor
A current is a stream of moving charges, so a current-carrying wire in a magnetic field feels the sum of the Lorentz forces on all its carriers (NCERT Section 4.2.3). Consider a straight rod of cross-sectional area $A$ and length $l$ carrying a steady current $I$, with mobile carriers of charge $q$, number density $n$ and average drift velocity $\mathbf{v}_d$. The total number of carriers is $nlA$, and the force on them is
$$\mathbf{F} = (nlA)\,q\,\mathbf{v}_d \times \mathbf{B} = [\,jAl\,]\times\mathbf{B} = I\,\mathbf{l}\times\mathbf{B}$$
where $\mathbf{l}$ is a vector of magnitude $l$ pointing along the direction of the current $I$ (the current itself is not a vector). The magnitude of this force is
$$F = BIl\sin\theta$$
with $\theta$ the angle between the current direction and $\mathbf{B}$. Here $\mathbf{B}$ is the external field, not the field produced by the rod itself. For a wire of arbitrary shape the Lorentz force is found by treating it as a collection of short straight strips and summing their contributions.
This result is the gateway to the rest of the chapter. Two parallel wires each sit in the other's field and so attract or repel — covered in force between parallel currents — and a current loop in a field experiences a net torque that drives the moving-coil galvanometer.
Lorentz Force at a Glance
- Lorentz law: $\mathbf{F} = q[\mathbf{E} + \mathbf{v}\times\mathbf{B}]$ — electric part $q\mathbf{E}$ plus magnetic part $q(\mathbf{v}\times\mathbf{B})$.
- Magnitude: $F = qvB\sin\theta$; maximum at $\theta = 90^\circ$, zero when $\mathbf{v}\parallel\mathbf{B}$ or the charge is at rest.
- Direction: perpendicular to both $\mathbf{v}$ and $\mathbf{B}$, by the right-hand (screw) rule; reversed for a negative charge.
- No work: the magnetic force is always $\perp \mathbf{v}$, so speed and kinetic energy stay constant — only direction changes.
- Unit: $\mathbf{B}$ is measured in tesla, $1\,\text{T} = 1\,\text{N/(A·m)}$; $1\,\text{gauss} = 10^{-4}\,\text{T}$.
- On a wire: $\mathbf{F} = I\,\mathbf{l}\times\mathbf{B}$, magnitude $BIl\sin\theta$, with $\mathbf{B}$ the external field.