# Magnetic Field of a Long Solenoid (Cylindrical Coil) Let's analyze the magnetic field everywhere (inside and outside) a very long solenoid (cylindrical coil) of: - Radius \( R \) - \( N \) turns per unit length - Steady current \( I \) Let’s use **Ampère’s Law** (the integral form of the Biot–Savart law is not as convenient due to symmetry, but Ampère's Law is a direct consequence and much easier here): \[ \oint \vec{B} \cdot d\vec{\ell} = \mu_ I_{\text{enc}} \] where \( I_{\text{enc}} \) is the total current enclosed by the loop. --- ## 1. **Field Inside the Solenoid (\( r < R \))** ### Symmetry - Choose an Amperian loop: a circle of radius \( r < R \) centered on the axis. - By symmetry, \( B \) is tangent to the loop and has the same magnitude everywhere on it. ### Apply Ampère’s Law \[ \oint \vec{B} \cdot d\vec{\ell} = B \cdot (2\pi r) \] The current enclosed is: \[ I_{\text{enc}} = (\text{number of turns per length}) \times (\text{length}) \times (\text{current per turn}) = N \cdot (\text{length of loop}) \cdot I \] But our Amperian loop runs parallel to the axis for the length, so for any length \( l \): \[ I_{\text{enc}} = N l I \] But along our circular loop, the length is just \( l \) (which cancels in both sides). So, \[ B (2\pi r) = \mu_ N l I \] But for unit length (\( l = 1 \)): \[ B (2\pi r) = \mu_ N I \] \[ B = \frac{\mu_ N I}{2\pi r} \] But this is **not correct** for inside the solenoid. In fact, the current enclosed is all the current through all wires passing through the circle of radius \( r \). For a tightly wound solenoid, **all the current is enclosed for \( r < R \)**, as all the turns pass through the interior. However, for an ideal solenoid (very long, tightly wound), the field inside is **uniform** and along the axis: \[ B = \mu_ N I \] **Summary:** - **Inside the solenoid (\( r < R \)), the magnetic field is uniform and given by:** \[ \boxed{B = \mu_ N I \qquad (r < R)} \] **Direction:** along the axis of the solenoid (right-hand rule with current direction). --- ## 2. **Field Outside the Solenoid (\( r > R \))** - Pick a circular Amperian loop of radius \( r > R \). - The net current passing through the loop is the same as the total current in all the turns per unit length times the length. But for every turn going "up" inside, there is a return current "down" outside (the wires loop back), and for an **ideal infinitely long solenoid**, these cancel, so: \[ I_{\text{enc}} = \] Thus, \[ \oint \vec{B} \cdot d\vec{\ell} = \implies B = \] **Summary:** - **Outside the solenoid (\( r > R \)), the magnetic field is zero:** \[ \boxed{B = \qquad (r > R)} \] --- ## 3. **Final Answers** ### **Magnetic Field Everywhere** \[ B(r) = \begin{cases} \mu_ N I, & \text{for } r < R \\ , & \text{for } r > R \end{cases} \] - **Inside (\( r < R \)):** Uniform, directed along the coil axis. - **Outside (\( r > R \)):** Zero (for infinitely long solenoid). --- ## **Visualization**  *Magnetic field lines are dense and uniform inside the solenoid and nearly zero outside for a very long solenoid.* --- ## **Note** - For a real (finite) solenoid, the field outside is not exactly zero, but for a "very long" solenoid, this is a very good approximation. - The Biot–Savart Law, in practice, leads to the same result but is much harder to use due to the solenoid's geometry. --- **References:** - Griffiths, *Introduction to Electrodynamics*, Sections 5.2.3–5.2.4 - Purcell & Morin, *Electricity and Magnetism*, Section 6.2