Magnetic Field Due to Different Geometries — Wire, Loop, Solenoid, Toroid Formulas

Question

What are the magnetic field expressions for a straight wire, circular loop, solenoid, and toroid — and when do we use each formula?

Solution — Step by Step

For a long straight wire carrying current $I$ , at perpendicular distance $r$ :

B = \frac{\mu_0 I}{2\pi r}

The field forms concentric circles around the wire (use right-hand thumb rule). This is the most frequently used formula — appears in nearly every magnetism problem set.

At the centre of a circular loop of radius $R$ carrying current $I$ :

B_{\text{centre}} = \frac{\mu_0 I}{2R}

On the axis at distance $x$ from the centre:

B_{\text{axis}} = \frac{\mu_0 IR^2}{2(R^2 + x^2)^{3/2}}

For $N$ turns, multiply by $N$ . Notice that when $x = 0$ , the axis formula reduces to the centre formula — a good sanity check.

Inside an ideal long solenoid with $n$ turns per unit length:

B_{\text{inside}} = \mu_0 nI

Outside the solenoid: $B \approx 0$ . The field inside is uniform and parallel to the axis — like a bar magnet but uniform throughout.

Inside the toroid (within the windings), for $N$ total turns and mean radius $r$ :

B = \frac{\mu_0 NI}{2\pi r}

Outside the toroid: $B = 0$ (both inside the hole and outside the ring). A toroid is essentially a solenoid bent into a circle — the field is confined entirely within the windings.

graph TD
    A[What geometry?] --> B{Straight wire?}
    B -->|Yes| C["B = mu0 I / 2pi r"]
    B -->|No| D{Circular loop?}
    D -->|Yes, at centre| E["B = mu0 I / 2R"]
    D -->|Yes, on axis| F["B = mu0 IR^2 / 2(R^2+x^2)^3/2"]
    D -->|No| G{Solenoid?}
    G -->|Yes| H["B = mu0 n I (inside)"]
    G -->|No| I{Toroid?}
    I -->|Yes| J["B = mu0 NI / 2pi r (inside)"]

Why This Works

All four formulas come from the same law — either Biot-Savart (for loops and finite wires) or Ampere’s circuital law (for solenoids, toroids, and infinite wires). The geometry determines which method is easier. Ampere’s law works beautifully when there is a clear symmetry that makes $\oint \vec{B} \cdot d\vec{l}$ tractable.

For JEE and NEET, memorise all four formulas cold. But also understand WHEN to use each: see a long wire? First formula. See a coil? Check if they want centre or axis point. See many turns wound on a cylinder? Solenoid. Wound in a ring? Toroid.

Alternative Method

For a finite straight wire of length $L$ , the field at perpendicular distance $d$ from the midpoint is:

B = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)

where $\alpha_1$ and $\alpha_2$ are the angles subtended by the two ends. For an infinite wire, both angles approach $90°$ , giving $B = \frac{\mu_0 I}{2\pi d}$ — consistent with Step 1.

Common Mistake

Students mix up $n$ (turns per unit length) and $N$ (total turns) in solenoid and toroid formulas. Solenoid uses $n = N/L$ ; toroid uses total $N$ . Writing $B = \mu_0 NI$ for a solenoid gives wrong dimensions. Always check: solenoid formula has no $2\pi$ in the denominator; toroid formula does.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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