Solenoid Magnetic Field Calculator

Find internal field B = μ₀nI for ideal solenoid

Parameters

turns/mⓘ
Aⓘ
Show Trail

Controls

xⓘ

Calculated Values

Magnetic Field:
0.00;T0.00;T

Examples

n=500, I=2 A

Lab solenoid.

    Visualization

    Magnetic Field of a Solenoid

    An ideal solenoid is a long helical coil with n turns per unit length carrying current I. Inside the mid-region (far from ends), the field is uniform, parallel to the axis, and given by B = μ₀nI. This is derived from superposing fields of many circular loops or by applying Ampère's law to a rectangular Amperian loop straddling the interior.

    Turn density n = N/L where N is total turns and L is solenoid length in meters. Example: 500 turns/m with I = 2 A gives B = (4π×10⁻⁷)(500)(2) ≈ 1.26×10⁻³ T ≈ 1.3 mT — strong enough to pick up paper clips with an iron core.

    Inserting a ferromagnetic core multiplies the field: B = μ₀μ_r nI where μ_r can exceed 1000 for soft iron at moderate fields. Electromagnets, relays, and doorbell ringer coils exploit this. Saturation limits B at high I when core domains align.

    Outside an ideal infinite solenoid, B ≈ 0. Real finite solenoids show fringing at ends where field lines bulge outward. A toroid (coil bent into a donut) confines all field lines inside the core with negligible external field — used in transformers and inductors.

    Solenoids are the magnetic analog of parallel-plate capacitors for uniform fields. MRI gradient coils, particle detector magnets, and lab demagnetizers all use solenoid geometry or stacks thereof.

    Class 12 compares solenoid and toroid fields; JEE may ask to find n or I given target B inside a given geometry.

    Key Concepts

    • B = μ₀nI (ideal interior)
    • n = N/L turns per meter
    • B ∝ n and I
    • Core: B = μ₀μ_r nI
    • Fringing at ends of finite coil
    • Toroid: field confined inside

    Real-World Applications

    • Electromagnets and relays
    • MRI gradient and shim coils
    • Inductors and transformers (toroid)
    • Demagnetizing coils
    • Class 12 solenoid numericals

    Explore Further

    More magnetism tools

    Physics Equations

    Solenoid:
    B=μ0nIB = \mu_0 n I

    Step-by-Step Solution

    See how the main results are calculated.

    1

    Step 1: Ideal Solenoid Model

    Equation:

    B=μ0nIB = \mu_0 n I

    Explanation:

    n = turns per unit length (turns/m); field uniform and parallel inside.

    2

    Step 2: Relate n to Total Turns

    Equation:

    n=NLn = \frac{N}{L}

    Explanation:

    N total turns over solenoid length L (m).

    3

    Step 3: Given

    Result:

    n=500 turns/m,I=2 An = 500\ \text{turns/m},\quad I = 2\ \text{A}
    4

    Step 4: Calculate B

    Calculation:

    B=1.2566e−6×500×2=1.2566e−3 TB = 1.2566e-6 \times 500 \times 2 = 1.2566e-3\ \text{T}

    Result:

    B=1.2566e−3TB = 1.2566e-3 T
    5

    Step 5: With Iron Core

    μ_r can be hundreds–thousands for soft iron.

    Equation:

    B=μ0μrnIB = \mu_0 \mu_r n I

    Explanation:

    Electromagnets use cores to multiply field strength.

    6

    Step 6: Applications

    Relays, MRI gradient coils, lab demagnetizers.

    Explanation:

    Outside ideal solenoid, B ≈ 0; ends show fringing.

    Frequently Asked Questions (FAQ)

    Short solenoid?

    End corrections needed; field weaker at ends.

    Toroid?

    B = μ₀nI confined inside core; no external field.

    Practice MCQs

    1. Ideal solenoid field:
    2. n is:
    3. Iron core:
    4. Double n and I:
    5. Outside ideal solenoid:
    6. Solenoid field direction: