Magnetic Fields from Currents: Straight and Solenoid (Basic)
Magnetic Field Due To A Current-carrying Conductor
Electric charges at rest create electric fields. However, when electric charges are in motion, they produce both electric fields and magnetic fields. An electric current, which is essentially a flow of electric charges, is a source of magnetic field in the space around it. This discovery was first made by Hans Christian Ørsted in 1820, who observed that a compass needle was deflected when placed near a current-carrying wire. This observation established a fundamental link between electricity and magnetism.
The pattern and strength of the magnetic field around a current-carrying conductor depend on the shape of the conductor. We will explore the magnetic fields produced by some simple configurations of current-carrying wires: a straight wire, a circular loop, and a solenoid.
Magnetic Field Due To A Current Through A Straight Conductor
Consider a long, straight wire carrying a steady electric current $I$. When you place compass needles around this wire, you observe that they align themselves in concentric circles around the wire. This indicates that the magnetic field lines around a straight current-carrying wire are concentric circles with the wire at their center, lying in a plane perpendicular to the wire.
Magnetic field lines around a straight current-carrying wire. The direction is given by the Right-Hand Thumb Rule.
The direction of the magnetic field lines reverses if the direction of the current is reversed.
The strength (magnitude) of the magnetic field ($\vec{B}$) at a point near the straight current-carrying wire depends on:
- The magnitude of the current ($I$) in the wire. The field is directly proportional to the current.
- The perpendicular distance ($r$) of the point from the wire. The field is inversely proportional to the distance.
For a long (ideally infinite) straight wire carrying current $I$, the magnitude of the magnetic field at a perpendicular distance $r$ from the wire is given by:
$ B = \frac{\mu_0 I}{2\pi r} $
Here, $\mu_0$ is the permeability of free space, a constant equal to $4\pi \times 10^{-7} \, T \cdot m/A$. The presence of a magnetic material near the wire would change this value (replace $\mu_0$ with the material's permeability $\mu$).
Right-hand Thumb Rule
The Right-Hand Thumb Rule is a convenient way to determine the direction of the magnetic field around a straight current-carrying wire.
Rule: Imagine grasping the current-carrying wire in your right hand such that your thumb points in the direction of the electric current ($I$). Then, your fingers will curl around the wire in the direction of the magnetic field lines ($\vec{B}$).
Applying the Right-Hand Thumb Rule. Thumb points in current direction, fingers curl in field direction.
For example, if the current flows upwards, the magnetic field lines form anti-clockwise circles when viewed from above. If the current flows into the page, the field lines form clockwise circles.
Magnetic Field Due To A Current Through A Circular Loop
Consider a circular loop of wire carrying a current $I$. To find the magnetic field produced by this loop, we can imagine it made up of many small straight current elements and use the Biot-Savart Law (as done in Section I2 of the previous topic). However, for a basic understanding, we focus on the pattern and the field strength at the center of the loop.
Magnetic field lines around a circular current loop. Field lines are dense at the center, becoming circular far away.
Near the wire segments of the loop, the field lines are almost circular, similar to a straight wire. As you move towards the center of the loop, the field lines from all the current elements add up. The field lines become straighter and are concentrated along the axis of the loop. At the exact center of the loop, the magnetic field is perpendicular to the plane of the loop.
The direction of the magnetic field at the center (and along the axis) is given by a variant of the Right-Hand Rule:
Right-Hand Rule for Circular Current: Curl the fingers of your right hand in the direction of the current flowing in the loop. Your extended thumb will point in the direction of the magnetic field at the center (and along the axis) of the loop.
Applying the Right-Hand Rule for a circular loop. Fingers curl in current direction, thumb points in field direction on axis.
The magnitude of the magnetic field at the center of a circular loop of radius $R$ carrying current $I$ is given by:
$ B_{center} = \frac{\mu_0 I}{2R} $
If the coil consists of $N$ closely spaced turns, each turn contributes independently to the field (assuming they are effectively at the same radius), so the total field at the center is $N$ times the field of a single loop:
$ B_{center} = \frac{\mu_0 N I}{2R} $
Magnetic Field Due To A Current In A Solenoid
A solenoid is a coil of wire wound in the shape of a cylinder. When an electric current flows through a solenoid, it produces a magnetic field. The magnetic field pattern of a solenoid is similar to that of a bar magnet.
Magnetic field lines of a solenoid carrying current, resembling a bar magnet.
For a long solenoid (where the length is much greater than the diameter), the magnetic field is:
- Nearly uniform and strong inside the solenoid, directed along its axis.
- Very weak and almost negligible outside the solenoid.
The direction of the magnetic field inside the solenoid is given by the Right-Hand Rule for circular currents, applied to the individual turns of the solenoid. If you curl the fingers of your right hand in the direction of the current in the windings, your extended thumb points in the direction of the magnetic field inside the solenoid. One end of the solenoid acts like a North pole and the other like a South pole.
The magnitude of the magnetic field inside a long solenoid depends on:
- The number of turns per unit length ($n$).
- The current ($I$) flowing through the solenoid.
The magnitude of the magnetic field inside a long solenoid is given by:
$ B = \mu_0 n I $ (inside a long solenoid)
Here, $n = N/L$, where $N$ is the total number of turns and $L$ is the length of the solenoid. $\mu_0$ is the permeability of free space.
This formula shows that the field inside a long solenoid is uniform (independent of position inside, except near the ends) and directly proportional to the number of turns per unit length and the current.
Solenoids are widely used to produce strong, controllable magnetic fields. They are the basis for electromagnets, used in various applications from lifting heavy iron objects to scientific experiments and medical imaging (MRI). Inserting a ferromagnetic material like soft iron into the core of a solenoid greatly increases the strength of the magnetic field, creating a powerful electromagnet.
Example 1. A long straight wire carries a current of 5 A. Calculate the magnitude of the magnetic field at a point 10 cm away from the wire. ($\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$).
Answer:
Given:
Current, $I = 5 \, A$
Distance from the wire, $r = 10 \, cm = 0.10 \, m$
Using the formula for the magnetic field due to a long straight wire:
$ B = \frac{\mu_0 I}{2\pi r} $
Substitute the values:
$ B = \frac{(4\pi \times 10^{-7} \, T \cdot m/A) \times (5 \, A)}{2\pi \times (0.10 \, m)} $
Cancel $2\pi$ from numerator and denominator:
$ B = \frac{2 \times 10^{-7} \times 5}{0.10} \, T $
$ B = \frac{10 \times 10^{-7}}{0.10} \, T = \frac{10^{-6}}{10^{-1}} \, T = 10^{-5} \, T $
The magnitude of the magnetic field at 10 cm from the wire is $10^{-5} \, T$.
Example 2. A circular coil has 100 turns and a radius of 5 cm. If a current of 0.5 A flows through the coil, calculate the magnetic field at the center of the coil.
Answer:
Given:
Number of turns, $N = 100$
Radius of the coil, $R = 5 \, cm = 0.05 \, m$
Current, $I = 0.5 \, A$
Using the formula for the magnetic field at the center of a circular coil:
$ B_{center} = \frac{\mu_0 N I}{2R} $
Substitute the values:
$ B_{center} = \frac{(4\pi \times 10^{-7} \, T \cdot m/A) \times 100 \times (0.5 \, A)}{2 \times (0.05 \, m)} $
$ B_{center} = \frac{4\pi \times 10^{-7} \times 50}{0.10} \, T $
$ B_{center} = \frac{200\pi \times 10^{-7}}{0.10} \, T = 2000\pi \times 10^{-7} \, T = 2\pi \times 10^{-4} \, T $
$ B_{center} \approx 2 \times 3.14159 \times 10^{-4} \, T \approx 6.283 \times 10^{-4} \, T $
The magnetic field at the center of the coil is approximately $6.283 \times 10^{-4} \, T$.