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What is the formula for the magnetic field due to a solenoid?

$\mu nI$

$\mu n^2$

$\mu I^2$

$\mu n^2 I^2$

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalMachine

Option A

$\mu nI$

Quick Summary:

The magnetic field inside a long, tightly wound, ideal solenoid carrying a current $I$ is uniform, parallel to its axis, and given by $B = \mu n I$ . Here, $\mu$ is the magnetic permeability of the core material ( $\mu = \mu_0 \mu_r$ ), $n$ is the number of turns per unit length, and $I$ is the electric current.

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$B = \mu n I$ — Magnetic field inside a long solenoid

$n = \frac{N}{L}$ — Turn density (turns per unit length)

$B_{end} = \frac{1}{2} \mu n I$ — Magnetic field strength at either end of a solenoid

⚙️ Working Principle

When an electric current flows through the turns of a solenoid, each helical turn creates a magnetic field loop. Inside the solenoid, the magnetic field vectors produced by adjacent turns add constructively along the central axis, resulting in a strong, uniform magnetic field, while outside the solenoid, the field components cancel out, making the external field nearly zero.

📌 Key Points

▸
The magnetic field inside an ideal, infinitely long solenoid is uniform and directed along the axis.
▸
The field strength depends directly on core permeability ( $\mu$ ), turn density ( $n$ ), and current ( $I$ ).
▸
The magnetic field strength at the extreme end of a long solenoid is exactly half of its value at the center.

✅ Advantages

▸
Produces a highly uniform and controllable magnetic field along its internal axis.
▸
Field magnitude can be easily controlled by adjusting the excitation current or turn density.

❌ Disadvantages / Limitations

▸
Magnetic flux leakage occurs at the ends, making the field non-uniform near edges.
▸
High currents lead to $I^2R$ resistive heating in the conductor windings.

🛠️ Applications / Uses

▸
Electromagnets, relays, and electromechanical actuators.
▸
Inductors and transformers in electrical power systems and machines.

📄 Additional Information

▸
In free space, $\mu = \mu_0 = 4\pi \times 10°{-7} \text{ H/m}$ , so $B = \mu_0 n I$ .
▸
Option B ( $\mu n^2$ ), Option C ( $\mu I^2$ ), and Option D ( $\mu n^2 I^2$ ) are mathematically incorrect and do not represent any magnetic field formula.

📊 Diagram / Illustration

✅

A is correct — The magnetic field inside an ideal long solenoid is given by $B = \mu n I$ , where $n$ is the number of turns per unit length.

Core Concepts Used

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Ampere's Circuital Law Solenoids and Electromagnets Magnetic Permeability

💡 EXAM TIP

Remember that at the precise physical end of a long solenoid, the field drops to $B_{end} = \frac{1}{2} \mu n I$ , a common trick question in GATE and AE/JE competitive exams.

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