Magnetic field due to different geometries — wire, loop, solenoid, toroid formulas

Question

Compare the magnetic field formulas for: (a) a long straight wire at distance $r$ , (b) a circular loop at its centre, (c) a solenoid, and (d) a toroid. For each, state the formula and when to use it. A solenoid has 500 turns per metre and carries 2 A. Find $B$ inside.

(CBSE 12 + NEET + JEE Main staple)

Solution — Step by Step

Geometry	Formula	Where it applies
Long straight wire	$B = \dfrac{\mu_0 I}{2\pi r}$	At perpendicular distance $r$ from wire
Circular loop (centre)	$B = \dfrac{\mu_0 I}{2R}$	At the centre of a single loop of radius $R$
Solenoid (inside)	$B = \mu_0 n I$	Inside a long solenoid, $n$ = turns per unit length
Toroid (inside)	$B = \mu_0 n I$	Inside the toroid, $n = N/(2\pi r)$

All four come from Ampere’s circuital law or Biot-Savart law. The geometry determines which formula applies.

Given: $n = 500$ turns/m, $I = 2$ A

B = \mu_0 n I = 4\pi \times 10^{-7} \times 500 \times 2

B = 4\pi \times 10^{-4} = \mathbf{1.26 \times 10^{-3} \text{ T}} \approx \mathbf{12.6 \text{ gauss}}

The field inside an ideal solenoid is uniform — it does not depend on position inside.

When you see a problem, identify the geometry first, then pick the formula. The flowchart below handles 90% of exam questions.

flowchart TD
    A["What is the current-carrying geometry?"] --> B{"Shape?"}
    B -->|"Straight wire"| C["B = μ₀I/(2πr)"]
    B -->|"Circular loop"| D{"Where is the point?"}
    B -->|"Solenoid"| E["B = μ₀nI (inside)"]
    B -->|"Toroid"| F["B = μ₀nI (inside ring)"]
    D -->|"At centre"| G["B = μ₀I/(2R)"]
    D -->|"On axis at distance x"| H["B = μ₀IR²/2(R²+x²)^(3/2)"]
    C --> I["Direction: right-hand thumb rule"]
    G --> I
    E --> I
    F --> I

Why This Works

All magnetic fields from steady currents ultimately come from the Biot-Savart law:

d\vec{B} = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l} \times \hat{r}}{r^2}

For a long straight wire, integrating this gives the $1/(2\pi r)$ dependence. For a loop at the centre, symmetry simplifies the integral to $\mu_0 I/(2R)$ . For a solenoid, Ampere’s law with a rectangular loop gives the clean result $B = \mu_0 nI$ inside.

The solenoid formula is particularly elegant — the field inside depends only on the current and how tightly the coils are wound (turns per metre), not on the radius or total length (as long as it is long enough).

Alternative Method — Ampere’s Law for Quick Derivation

For the solenoid, draw a rectangular Amperian loop with one side inside and one side outside. The field outside an ideal solenoid is zero. Applying $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$ :

B \cdot l = \mu_0 \cdot (nl) \cdot I

B = \mu_0 nI

For NEET, memorise the four formulas directly — derivations are rarely asked. For JEE Main, you also need the on-axis formula for a loop: $B = \dfrac{\mu_0 IR^2}{2(R^2 + x^2)^{3/2}}$ . This reduces to $\mu_0 I/(2R)$ at $x = 0$ (centre) and to $\mu_0 IR^2/(2x^3)$ far away ( $x \gg R$ ), which looks like a magnetic dipole.

Common Mistake

The solenoid formula uses $n$ = turns per unit LENGTH (turns/m), not total number of turns $N$ . If a problem says “a solenoid of length 50 cm with 1000 turns,” then $n = 1000/0.5 = 2000$ turns/m. Using $N = 1000$ directly in $B = \mu_0 n I$ gives a wrong answer that is off by orders of magnitude. Always convert to turns per metre first.

Question

Solution — Step by Step

Why This Works

Alternative Method — Ampere’s Law for Quick Derivation

Common Mistake

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