Magnetic Force

Where the chapter begins: Oersted's observation

• A current in a straight wire deflects a nearby compass needle; the needle sets itself tangential to a circle centred on the wire, in a plane perpendicular to it. Reversing the current reverses the deflection, and iron filings sprinkled round the wire settle into concentric circles.
• Conclusion: moving charges (currents) produce a magnetic field in the space around them. Everything in this chapter follows from that plus the Lorentz force.
• Convention: a dot is a current or field out of the page, a cross is one into the page.

Sources and fields

• A charge $Q$ produces an electric field $\vec{E}$; another charge $q$ placed there feels $\vec{F} = q\vec{E}$. Currents and moving charges produce, in addition, a magnetic field $\vec{B}(\vec{r})$.
• $\vec{B}$ is a vector field defined at every point, and it obeys the principle of superposition: the field of several sources is the vector sum of the individual fields. This chapter treats fields that do not vary with time.

The Lorentz force

• A charge $q$ at position $\vec{r}$ moving with velocity $\vec{v}$ in both fields feels $$\vec{F} = q\left[\vec{E}(\vec{r}) + \vec{v}\times\vec{B}(\vec{r})\right] = \vec{F}_{\text{electric}} + \vec{F}_{\text{magnetic}}$$
• The magnetic part alone, the answer CBSE wants for the one-mark question: $$\vec{F} = q\left(\vec{v}\times\vec{B}\right), \qquad F = qvB\sin\theta$$ with $\theta$ the angle between $\vec{v}$ and $\vec{B}$.
• Direction: perpendicular to the plane containing $\vec{v}$ and $\vec{B}$, given by the right-hand screw rule for the cross product (equivalently by the right-hand rule). Always state this — the direction is half the mark.

Features to quote

• It depends on $q$, $v$ and $B$; the force on a negative charge is opposite to that on a positive charge moving the same way.
• It vanishes if $\vec{v}$ is parallel or antiparallel to $\vec{B}$, since the cross product is then zero.
• It vanishes if the charge is at rest: only a moving charge feels a magnetic force.
• It acts sideways, perpendicular to both $\vec{v}$ and $\vec{B}$ — hence it can never change the speed.

Definition of the tesla

• From $F = qvB\sin\theta$, put $q = 1\ \text{C}$, $v = 1\ \text{m s}^{-1}$, $\theta = 90^\circ$: $$B = \dfrac{F}{qv}, \qquad 1\ \text{T} = \dfrac{1\ \text{N}}{(1\ \text{C})(1\ \text{m s}^{-1})}$$
• One tesla is the magnitude of magnetic field which exerts a force of one newton on a charge of one coulomb moving perpendicular to the field at one metre per second.
• Dimensionally $[B] = [F/qv] = [\text{M T}^{-2}\text{A}^{-1}]$. Non-SI unit: $1\ \text{gauss} = 10^{-4}\ \text{T}$; the earth's field is about $3.6\times10^{-5}\ \text{T}$, so the tesla is a large unit.

Force on a current-carrying conductor — the derivation CBSE sets

• Take a straight rod of length $l$ and uniform cross-section $A$, carrying a steady current $I$, in an external field $\vec{B}$.
• Let $n$ be the number density of mobile carriers, each of charge $q$, drifting with average velocity $\vec{v}_d$. The rod holds $nlA$ carriers, so the total force is $$\vec{F} = (nlA)\,q\,\vec{v}_d\times\vec{B}$$
• But $nq\vec{v}_d = \vec{j}$, the current density, and $\lvert nq\vec{v}_d\rvert A = I$, the current. Hence $$\vec{F} = \left[jAl\right]\times\vec{B} = I\left(\vec{l}\times\vec{B}\right), \qquad F = BIl\sin\theta$$
• $\vec{l}$ has magnitude $l$ and points along the current; the current $I$ itself is a scalar — the vector sign is transferred from $\vec{j}$ to $\vec{l}$.
• $\vec{B}$ here is the external field, not the field the rod itself produces.
• For a wire of arbitrary shape, treat it as a collection of strips $d\vec{l}$ and integrate: $\vec{F} = I\displaystyle\int d\vec{l}\times\vec{B}$.
• The force is maximum ($BIl$) when the conductor is perpendicular to $\vec{B}$, and zero when it lies along $\vec{B}$.

Standard problem: a wire suspended in mid-air

• A wire of mass $m$ and length $l$ carrying current $I$ floats in a horizontal field, or hangs from strings that have gone slack. Equating the upward magnetic force to the weight: $$mg = BIl \;\Rightarrow\; B = \dfrac{mg}{Il}, \qquad I = \dfrac{mg}{lB}$$
• Worked case from the NCERT text and from Foreign 2015: $m = 200\ \text{g} = 0.2\ \text{kg}$, $l = 1.5\ \text{m}$, $I = 2\ \text{A}$, so $B = \dfrac{0.2\times9.8}{2\times1.5} = 0.65\ \text{T}$.
• Note that only $m/l$, the mass per unit length, is really needed. The earth's field ($\sim 4\times10^{-5}\ \text{T}$) is ignored.

Diagram in words

• Draw $\vec{v}$ as a horizontal arrow and $\vec{B}$ at an angle $\theta$ to it in the same plane, with a shaded plane containing both. The force $\vec{F}$ is a third arrow perpendicular to that plane, pointing out of it for a positive charge and into it for a negative one — mark the two oppositely for $+q$ and $-q$.
• For the current-carrying rod: rod horizontal, current arrow along it, field arrows across it, force arrow perpendicular to both; label $l$ along the rod in the direction of $I$.

Results to carry away

• $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$; magnetic part $q(\vec{v}\times\vec{B})$, magnitude $qvB\sin\theta$.
• $\vec{F} = I(\vec{l}\times\vec{B})$, magnitude $BIl\sin\theta$, maximum at $90^\circ$, zero at $0^\circ$.
• $B = F/(qv)$ defines the tesla; $1\ \text{G} = 10^{-4}\ \text{T}$.
• Suspension condition $BIl = mg$.