Sublevo
ISC 2027
All chaptersPhysics · Unit 3

Magnetism

7 articles31 formulas39 ways the board asks it
PHYMagnetic Field & Forces

Biot–Savart Law & Field due to Currents

The Biot–Savart law gives the magnetic field produced by a current element, and integrating it (or using Ampere's law as a shortcut) yields the standard results for a straight wire, a circular loop, a point on the loop's axis, a solenoid and a toroid. ISC numericals are almost always direct substitution into one of these standard formulas, so recognising the geometry is the whole skill.

Note that the field always wraps around the current (right-hand rule) and falls off with distance.

Biot–Savart law (magnitude)
dB=μ04π I dl sin⁡θr2dB = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,dl\,\sin\theta}{r^2}
dldl current-element length, θ\theta angle between the element and r⃗\vec{r}, rr distance to the field point, μ0/4π=10−7 T m A−1\mu_0/4\pi = 10^{-7}\ \mathrm{T\,m\,A^{-1}}.
Long straight wire
B=μ0I2πaB = \dfrac{\mu_0 I}{2\pi a}
aa perpendicular distance from the wire; valid for an infinitely long straight conductor.
Centre of a circular coil (NN turns)
B=μ0NI2RB = \dfrac{\mu_0 N I}{2R}
RR radius of coil, NN number of turns, field directed along the axis through the centre.
Axial field of a circular coil
B=μ0NIR22 (R2+x2)3/2B = \dfrac{\mu_0 N I R^2}{2\,(R^2 + x^2)^{3/2}}
xx distance of the point from the centre along the axis; reduces to μ0NI/2R\mu_0 NI/2R when x=0x=0.
Inside a long solenoid
B=μ0nIB = \mu_0 n I
n=N/Ln = N/L turns per unit length; field is uniform along the axis inside the solenoid.
Inside a toroid
B=μ0NI2πrB = \dfrac{\mu_0 N I}{2\pi r}
NN total turns, rr mean radius of the toroid; field is confined to the core and zero outside.
  • Use μ0/4π=10−7 T m A−1\mu_0/4\pi = 10^{-7}\ \mathrm{T\,m\,A^{-1}} in Biot–Savart; use μ0=4π×10−7\mu_0 = 4\pi\times10^{-7} in the loop/solenoid formulas.
  • For a circular coil always include the number of turns NN; a single loop is the N=1N=1 case.
  • The solenoid result B=μ0nIB=\mu_0 nI uses n=N/Ln=N/L (turns per metre), not the total turns NN.
  • The toroid uses the mean radius rr and total turns NN; the field outside both the core and the toroid is zero.
  • All lengths must be converted to metres (cm→×10−2 m\mathrm{cm}\to\times10^{-2}\ \mathrm{m}) before substituting.
  • Fields from several wires add as vectors; if two fields are mutually perpendicular the resultant is B12+B22\sqrt{B_1^2+B_2^2}.
  • The straight-wire field circles the conductor; direction is given by the right-hand grip rule (thumb along II, fingers along B⃗\vec{B}).
Where the marks go
  • Dropping the factor NN for a multi-turn coil, giving an answer NN times too small.
  • Using μ0I/2πa\mu_0 I/2\pi a (straight wire) for the centre of a loop, or vice versa — the loop has no π\pi in the denominator.
  • Forgetting to convert centimetres to metres, which throws the magnitude off by powers of ten.
  • In the solenoid, plugging in total turns NN instead of n=N/Ln=N/L turns per unit length.
How the board asks it
  • Numericalthe loop and solenoid formulas
    A circular coil of 200200 turns and radius 5 cm5\ \mathrm{cm} carries a current of 2.5 A2.5\ \mathrm{A}. Calculate the magnitude of the magnetic field at the centre of the coil. (μ0=4π×10−7 T m A−1)(\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}})
  • Derive / provethe biot–savart law
    Using the Biot–Savart law, derive an expression for the magnetic field at the centre of a circular current-carrying coil of radius rr having NN turns and carrying a current II.
  • Numericalvector addition of fields
    Two long straight parallel wires, 10 cm10\ \mathrm{cm} apart, carry currents of 5 A5\ \mathrm{A} and 7 A7\ \mathrm{A} in the same direction. Find the magnitude of the resultant magnetic field at a point midway between the two wires. (μ0/4π=10−7 T m A−1)(\mu_0/4\pi = 10^{-7}\ \mathrm{T\,m\,A^{-1}})
  • Define / statethe biot–savart law statement
    State the Biot–Savart law and write the expression for the magnitude of the magnetic field dBdB due to a small current element I dl⃗I\,d\vec{l} at a point distant rr from it.
  • Give reasonsfield inside vs outside a toroid
    Give a reason why the magnetic field at any point outside an ideal toroid is zero, while inside the core it is B=μ0NI/(2πr)B=\mu_0 N I/(2\pi r), where rr is the mean radius.

Practise this topic

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.