Magnetic Effects of Current
SRMJEEE – Physics – Chapter 2.2
1. Simple Explanation
When an electric current flows through a conductor, it generates a magnetic field around it.
- The field lines form closed loops that start from the north side of the current and return to the south side.
- The direction of the magnetic field produced by a current element is given by the right‑hand rule: point your thumb in the direction of the current, your fingers curl in the direction of the magnetic field.
The magnetic field produced by a current can exert forces on other currents or on moving charges (the Lorentz force). These effects are the basis of electromagnets, motors, generators, transformers, and many other devices.
2. Key Points & Formulae
| Concept | Formula (SI) | Description |
|---|---|---|
| Biot–Savart Law (magnetic field due to a current element) | (\displaystyle \mathbf{B}=\frac{\mu_0}{4\pi}\int \frac{I,d\mathbf{l}\times\hat r}{r^{2}}) | Gives the differential contribution of a current element (I,d\mathbf{l}) to the magnetic field at a field point. |
| Magnetic field of an infinite straight wire | (\displaystyle B=\frac{\mu_0 I}{2\pi r}) | Circular field lines around the wire; (r) is the perpendicular distance. |
| Magnetic field of a finite straight wire (length (2L)) | (\displaystyle B=\frac{\mu_0 I}{4\pi r}\left(\sin\theta_1+\sin\theta_2\right)) | (\theta_1,\theta_2) are the angles made by the wire ends with the field point. |
| Ampère’s Law (integral form) | (\displaystyle \oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\text{enc}}) | Useful for symmetrical situations (solenoids, toroids). |
| Magnetic field inside a long solenoid | (\displaystyle B=\mu_0 n I) | (n) = turns per unit length, (I) = current. |
| Magnetic field inside a toroid | (\displaystyle B=\frac{\mu_0 N I}{2\pi r}) | (N) = total turns, (r) = radial distance from centre. |
| Force between two parallel currents | (\displaystyle F/L=\frac{\mu_0 I_1 I_2}{2\pi d}) | (L) = length of the segment, (d) = separation. |
| Lorentz Force on a moving charge | (\displaystyle \mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})) | For purely magnetic case: (\mathbf{F}=q\mathbf{v}\times\mathbf{B}). |
| Magnetic force on a current‑carrying wire | (\displaystyle \mathbf{F}=I,\mathbf{L}\times\mathbf{B}) | (\mathbf{L}) is the vector length of the wire segment. |
| Magnetic torque on a current loop | (\displaystyle \boldsymbol{\tau}=I,\mathbf{A}\times\mathbf{B}) | (\mathbf{A}) = area vector of the loop. |
| Magnetic pressure in a solenoid | (\displaystyle P=\frac{B^{2}}{2\mu_0}) | Pressure exerted by the magnetic field. |
Notes
( \mu_0 = 4\pi\times10^{-7},\text{H/m}) (permeability of free space).
All equations assume non‑relativistic speeds and static fields.
3. Worked Example
Problem:
A long straight wire of radius (a=1,\text{mm}) carries a current (I=5,\text{A}).
(a) Find the magnetic field at a distance (r=5,\text{mm}) from the centre of the wire.
(b) Determine the force per unit length between this wire and a second identical wire placed (d=20,\text{mm}) apart, running parallel to the first.
(a) Magnetic field at (r=5\text{ mm})
The wire is long enough that the field can be treated as that of an infinite straight wire.
[ B=\frac{\mu_0 I}{2\pi r} =\frac{4\pi\times10^{-7},\text{H/m};\times 5,\text{A}} {2\pi\times 5\times10^{-3},\text{m}} =\frac{2\times10^{-6},\text{H}\cdot\text{A}/\text{m}} {5\times10^{-3},\text{m}} =4.0\times10^{-4},\text{T};(\text{or};0.4,\text{mT}) ]
Answer: (B = 0.4\ \text{mT}) at (r=5\ \text{mm}).
(b) Force per unit length between the two wires
Using the