text: # Current-carrying Wires and Solenoids
  
  This topic is part of the HSC Physics course under the section Magnetism.
  
  ### HSC Physics Syllabus
  
  • use magnetic field lines to model qualitatively the direction and strength of magnetic fields produced by magnets, current-carrying wires and solenoids and relate these fields to their effect on magnetic materials that are placed within them (ACSPH083)
  • conduct investigations into and describe quantitatively the magnetic fields produced by wires and solenoids, including: (ACSPH106, ACSPH107)
  – B = \frac{\mu_0 I}{2\pi r}
  – B = \frac{\mu_0 NI}{L}
  
  ### Magnetic Field Around Current-carrying Conductors
  
  When an electric current flows through a wire, it generates a magnetic field that circles the wire. This magnetic field can be visualised using magnetic field lines that wrap around the wire in concentric circles.
  
  The generation of a magnetic field by an electric current is one of the foundational principles of electromagnetism, discovered by Hans Christian Ørsted in the 19th century.
  
  When Ørsted first observed that a compass needle was deflected by a nearby electric current, he demonstrated that moving charges (the flow of electrons in the wire) produce a magnetic field that affects other magnetic materials, like the needle of a compass.
  
  When electrons move through a conductor, they create a magnetic field that interacts with the "magnetic moments" of the electrons in nearby magnetic materials, aligning them with the field. This effect is much stronger in materials with domains, such as ferromagnetic materials, because the domains align to strengthen the overall magnetic field.
  
  A useful tool for determining the direction of the magnetic field is the right-hand rule. If you point the thumb of your right hand in the direction of the current, your fingers will curl in the direction of the magnetic field lines.
  
  The strength of the magnetic field is given by the equation:
  
  $$B = \frac{\mu_0 I}{2\pi r}$$
  
  where
  
  • B is the magnetic field strength or field density measured in Teslas (T)
  • \mu_0 is the magnetic permeability constant of free space = 4 \pi xx 10^{-7} T m A–1
  • I is the magnitude of current in the wire in Amperes (A)
  • r is the perpendicular distance from the wire in metres (m)
  
  ### Example 1
  
  Calculate the magnetic field strength 5 cm from a straight wire that is carrying 2.5 A current.
  
  Solution:
  
  $$B = \frac{\mu_0 I}{2\pi r}$$
  
  $$B = \frac{(4 \pi \times 10^{-7})(2.5)}{2\pi (0.05)}$$
  
  $$B = 1 \times 10^{-5} \text{ T}$$
  
  ### Magnetic Field of a Solenoid
  
  A solenoid is a fundamental component in the field of electromagnetism, commonly used in physics and engineering. It consists of a coil of wire, often wound tightly around a metallic core, which produces a uniform magnetic field when an electric current passes through it.
  
  • Coil: A solenoid is typically made by coiling insulated wire around a cylindrical form. The wire turns are usually wound closely to ensure a uniform magnetic field.
  
  • Magnetic Field: When current flows through the wire, it creates a magnetic field with distinct north and south poles. Inside the coil, the magnetic field lines are nearly parallel to each other, indicating a uniform field.
  
  • Core: While solenoids can function without a core, inserting a ferromagnetic core (e.g. iron) inside the coil greatly enhances the magnetic field's strength because the core becomes magnetised.
  
  The magnetic field inside a long solenoid is uniform and its strength can be calculated using the formula:
  
  $$B = \frac{\mu_0 N I}{L}$$
  
  where:
  
  • B is the magnetic field strength in Teslas (T)
  • \mu_0 is the magnetic permeability of free space
  • N is the total number of turns in the coil
  • L is the length of the solenoid in metres (m)
  • I is the magnitude of current through the solenoid in Amperes (A)
  
  The right-hand rule for solenoids helps to determine the direction of the magnetic field inside the solenoid when an electric current is passing through the coil. The way the right-hand rule used for solenoids is different to straight conductors or wires.
  
  For a solenoid, you can use the right-hand grip rule as follows:
  
  • Curl Your Fingers: Point the fingers of your right hand in the direction of the conventional current (from positive to negative) in the coils of the solenoid.
  
  • Thumb Points to North Pole: Your thumb, which is extended perpendicular to your fingers, will point towards the north pole of the solenoid's magnetic field.
  
  A solenoid becomes an electromagnet when it's wound around a ferromagnetic core, like iron. Here's how they are related:
  
  1. Magnetic Field Enhancement: The ferromagnetic core within the solenoid enhances the magnetic field created by the coil. The core's magnetic permeability is much greater than air or vacuum, so it concentrates the magnetic field lines within it, resulting in a stronger magnetic field.
  
  2. Consistent Field Direction: The solenoid's structure ensures that the magnetic field inside is uniform and directed along the axis of the coil. This creates a defined north and south pole, similar to a bar magnet, which is a characteristic of an electromagnet.
  
  3. Controlled Magnetism: An electromagnet, such as a solenoid with a core, can be easily controlled by starting or stopping the current, or by reversing its direction. This makes electromagnets incredibly versatile for applications where a controllable magnetic field is required.
  
  4. Uses: Electromagnets are widely used in technology, from electric motors and loudspeakers to magnetic locks and hard disk drives. A solenoid acts as an electromagnet when it's part of a circuit designed to convert electrical energy into mechanical work; for example, in solenoid valves, relays, or actuators.
  
  ### Example 2
  
  A solenoid is wound with 1000 turns per metre. The current in the solenoid is 20 A. What is the maximum magnetic field inside the solenoid measured in mT?
  
  Solution:
  
  $$B = \frac{\mu_0 N I}{L}$$
  
  $$B = (4\pi \times 10^{-7})(1000)(20)$$
  
  $$B = 0.025 \text{ T} = 25 \text{ mT}$$