Magnetism Formula

Magnetism is a fundamental force of nature intimately tied to the motion of electric charges. This force manifests itself in the form of attractive or repulsive interactions between objects, which is one of the key components of electromagnetism. Electromagnetism is one of the four fundamental forces of physics, the others being gravity, the weak nuclear force, and the strong nuclear force. Learn more about magnetism and its formula in this article by Extramarks

Definition of Magnetism

Magnetism is a fundamental physical phenomenon produced by the motion of electric charges, resulting in attractive and repulsive forces between objects. It is one aspect of the combined electromagnetic force. Here are the key points about magnetism:

1. Magnetic Fields and Forces: Magnetism is associated with magnetic fields, which exert forces on moving electric charges and magnetic dipoles (such as bar magnets). The region around a magnetic material or a moving electric charge where the magnetic force can be detected is called the magnetic field.
2. Magnetic Poles: Magnetic objects have two poles, commonly referred to as the north and south poles. Like poles repel each other, while opposite poles attract. These poles are where the magnetic field is strongest.

Magnetism Formula

The study of magnetism involves several formulas that describe different aspects of magnetic fields, forces, and interactions. Here are some key formulas related to magnetism:

Magnetic Force on a Moving Charge:

\[\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\]

• \(\mathbf{F}\) is the magnetic force.
• \(q\) is the electric charge.
• \(\mathbf{v}\) is the velocity of the charge.
• \(\mathbf{B}\) is the magnetic field.
• \(\times\) denotes the cross product, meaning the force is perpendicular to both the velocity and the magnetic field.

Magnetic Force on a Current-Carrying Wire:

\[\mathbf{F} = I(\mathbf{L} \times \mathbf{B})\]

• \(\mathbf{F}\) is the magnetic force.
• \(I\) is the current.
• \(\mathbf{L}\) is the length vector of the wire segment.
• \(\mathbf{B}\) is the magnetic field.

Biot-Savart Law

It is used to calculate the magnetic field created by a current element.

\[d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{L} \times \mathbf{\hat{r}}}{r^2}\]

• \(d\mathbf{B}\) is the infinitesimal magnetic field.
• \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\)).
• \(I\) is the current.
• \(d\mathbf{L}\) is the infinitesimal length of the wire.
• \(\mathbf{\hat{r}}\) is the unit vector from the current element to the point of observation.
• \(r\) is the distance from the current element to the point of observation.

Ampere’s Law

 It relates the integrated magnetic field around a closed loop to the electric current passing through the loop

\[\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}\]

• \(\oint \mathbf{B} \cdot d\mathbf{l}\) is the line integral of the magnetic field around a closed path.
• \(\mu_0\) is the permeability of free space.
• \(I_{\text{enc}}\) is the current enclosed by the path.

Magnetic Flux

\[\Phi_B = \int \mathbf{B} \cdot d\mathbf{A}\]

• \(\Phi_B\) is the magnetic flux.
• \(\mathbf{B}\) is the magnetic field.
• \(d\mathbf{A}\) is the differential area element perpendicular to the magnetic field.

Faraday’s Law of Electromagnetic Induction:

\[\mathcal{E} = -\frac{d\Phi_B}{dt}\]

• \(\mathcal{E}\) is the electromotive force (emf).
• \(\Phi_B\) is the magnetic flux.
• \(t\) is time.

The negative sign indicates the direction of the induced emf (Lenz’s Law).

These formulas are fundamental in understanding and calculating various aspects of magnetism in physics.

Solved Examples on Magnetism Formula

Example 1: A proton (\(q = 1.6 \times 10^{19} \, \text{C}\)) is moving with a velocity of \(5 \times 10^6 \, \text{m/s}\) perpendicular to a magnetic field of \(0.1 \, \text{T}\). Calculate the magnetic force acting on the proton.

Solution:

The magnetic force \(\mathbf{F}\) on a moving charge is given by:

\[\mathbf{F} = q (\mathbf{v} \times \mathbf{B})\]

Since the velocity is perpendicular to the magnetic field, the magnitude of the force is:

\[F = q v B \sin(\theta)\]

where \(\theta = 90^\circ\) and \(\sin(90^\circ) = 1\).

So,

\[F = (1.6 \times 10^{19} \, \text{C}) (5 \times 10^6 \, \text{m/s}) (0.1 \, \text{T})\]

\[F = 8 \times 10^{14} \, \text{N}\]

Answer: The magnetic force acting on the proton is \(8 \times 10^{14} \, \text{N}\).

Example 2: A solenoid has 500 turns per meter and carries a current of \(3 \, \text{A}\). Calculate the magnetic field inside the solenoid.

Solution:

The magnetic field \(\mathbf{B}\) inside a solenoid is given by:

\[B = \mu_0 n I\]

where:

\(\mu_0 = 4\pi \times 10^{7} \, \text{T m/A}\) (permeability of free space)

\(n = 500 \, \text{turns/m}\)

\(I = 3 \, \text{A}\)

So,

\[B = (4\pi \times 10^{7} \, \text{T m/A}) (500 \, \text{turns/m}) (3 \, \text{A})\]

\[B = 6\pi \times 10^{4} \, \text{T}\]

\[B \approx 1.88 \times 10^{3} \, \text{T}\]

Answer: The magnetic field inside the solenoid is approximately \(1.88 \times 10^{3} \, \text{T}\).

Example 3: A circular loop of radius \(0.2 \, \text{m}\) carries a current of \(10 \, \text{A}\). Calculate the magnetic field at the center of the loop.

Solution:

The magnetic field \(\mathbf{B}\) at the center of a circular loop is given by:

\[B = \frac{\mu_0 I}{2 R}\]

where:

\(\mu_0 = 4\pi \times 10^{7} \, \text{T m/A}\)

\(I = 10 \, \text{A}\)

\(R = 0.2 \, \text{m}\)

So,

\[B = \frac{4\pi \times 10^{7} \, \text{T m/A} \times 10 \, \text{A}}{2 \times 0.2 \, \text{m}}\]

\[B = \frac{4\pi \times 10^{6} \, \text{T m}}{0.4 \, \text{m}}\]

\[B = \pi \times 10^{5} \, \text{T}\]

\[B \approx 3.14 \times 10^{5} \, \text{T}\]

Answer: The magnetic field at the center of the loop is approximately \(3.14 \times 10^{5} \, \text{T}\).