When an electric current flows through a conductor it produces a magnetic field in the surrounding space.
The magnetic field can exert forces on other currents or on moving charges. These ideas form the core of Cambridge International AS & A Level Physics (Topic 20 – Magnetic fields) and are the basis for later topics such as electromagnetic induction, motors, and the definition of the ampere.
| Quantity | Symbol | Unit | Physical meaning |
|---|---|---|---|
| Magnetic flux density | \( \mathbf{B} \) | T (tesla) | Force per unit current per unit length; includes the effect of the material. |
| Magnetic field intensity | \( \mathbf{H} \) | A·m\(^{-1}\) | Magnetising force produced by currents; independent of the material. |
The two are related by the magnetic permeability \( \mu \):
\[
\mathbf{B}= \mu \mathbf{H},\qquad \mu = \mu{0}\mu{r},
\]
where \( \mu{0}=4\pi\times10^{-7}\;\text{T·m·A}^{-1} \) is the permeability of free space and \( \mu{r} \) is the relative permeability of the material.
Effect of magnetic materials
From the Biot–Savart law, the magnitude of the field at a perpendicular distance \( r \) from a long straight wire carrying current \( I \) is
\[
B = \frac{\mu_{0} I}{2\pi r}.
\]
Direction: right‑hand grip rule – thumb points in the direction of conventional current, fingers curl in the direction of \( \mathbf{B} \).
For a single loop carrying current \( I \), the field on the axis a distance \( x \) from the centre is
\[
B{\text{axis}} = \frac{\mu{0} I R^{2}}{2\,(R^{2}+x^{2})^{3/2}}.
\]
At the centre (\( x=0 \)) this reduces to
\[
B{\text{centre}} = \frac{\mu{0} I}{2R}.
\]
Direction: curl the fingers in the sense of the current; the thumb points along the field inside the loop.
Inside a long solenoid the field is almost uniform and parallel to the axis:
\[
B{\text{inside}} = \mu{0} n I.
\]
Outside an ideal (infinitely long) solenoid the field is essentially zero. The direction follows the same right‑hand rule as for a single loop.
\[
\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}.
\]
Consider a small segment of wire of length \( \mathrm{d}\mathbf{L} \) containing charge carriers of density \( n \) (charges per unit volume) moving with drift speed \( \mathbf{v}_{d} \). The number of carriers in the segment is \( nA\,\mathrm{d}L \) (where \( A \) is the cross‑sectional area) and the total charge is \( q = n e A\,\mathrm{d}L \).
Each carrier experiences the Lorentz force \( \mathrm{d}\mathbf{F}=q\,\mathbf{v}_{d}\times\mathbf{B} \). Summing over the segment gives
\[
\mathrm{d}\mathbf{F}= (n e A\,\mathrm{d}L)\,\mathbf{v}_{d}\times\mathbf{B}.
\]
But the current is \( I = n e A v_{d} \). Substituting,
\[
\mathrm{d}\mathbf{F}= I\,\mathrm{d}\mathbf{L}\times\mathbf{B}.
\]
Integrating over the whole conductor of length \( \mathbf{L} \) yields the familiar expression
\[
\boxed{\mathbf{F}= I\,\mathbf{L}\times\mathbf{B}}.
\]
Hence the magnitude is \( F = I L B \sin\theta \) and the direction follows the right‑hand rule (thumb → \( I\mathbf{L} \), fingers → \( \mathbf{B} \), palm → \( \mathbf{F} \)).
\[
B{1}= \frac{\mu{0} I_{1}}{2\pi d},
\]
where \( d \) is the centre‑to‑centre separation.
\[
F{21}= I{2} L B{1}= \frac{\mu{0} I{1} I{2} L}{2\pi d},
\]
directed by the right‑hand rule for \( \mathbf{L}{2}\times\mathbf{B}{1} \).
| Current directions | Resulting force |
|---|---|
| Both currents in the same direction (parallel) | Attractive – the wires pull toward each other. |
| Currents opposite (antiparallel) | Repulsive – the wires push away from each other. |
Two long wires are 4 cm apart and each carries 5 A in the same direction. Find the force per metre of length.
\[
\frac{F}{L}= \frac{\mu{0} I{1} I_{2}}{2\pi d}
= \frac{4\pi\times10^{-7}\times5\times5}{2\pi\times0.04}
= 1.25\times10^{-4}\;\text{N m}^{-1}.
\]
Thus the wires attract each other with a force of \(0.125\;\text{mN}\) per metre of length.
One ampere is that constant current which, if maintained in two straight, parallel conductors of infinite length and negligible circular cross‑section, placed 1 m apart in vacuum, would produce a force of \(2\times10^{-7}\;\text{N}\) per metre of length on each conductor.
This definition follows directly from the formula \(F/L = \mu{0}I^{2}/(2\pi d)\) with \(I{1}=I_{2}=1\;\text{A}\) and \(d=1\;\text{m}\).
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