Define Magnetic Flux it's Units and Dimensions , Expression for magnetic flux associated with variable magnetic field.

Define Magnetic Flux it's Units and Dimensions , Expression for magnetic flux associated with variable magnetic field.

Define magnetic flux. Give units and dimensions of magnetic flux. Give an expression for magnetic flux associated with variable magnetic field.

Magnetic flux ($\phi$) through any surface is defined as " the total number of magnetic lines passing through that surface."

Magnetic Flux

Consider a small surface of area A. Let $\hat{n}$ be the unit vector which is drawn normal to the plane of the surface.

If $\theta$ is the angle between $\hat{n}$ and the uniform magnetic field $\vec{B}$ , then the magnetic flux ($\phi$) through the surface is given by,

$\phi_B = \vec{B} \cdot \vec{A} = BA \cos\theta \tag{i}$

or 

$\phi_B = (B \cos\theta) A \tag{ii}$

Now $B \cos \theta$ is the component of the magnetic field normal to the plane of the surface

Then eqn. (ii) can be written as, 

$\Phi_B = BA \tag{iii}$

Thus, magnetic flux through a given surface is defined as the product of the area of the surface and the component of the magnetic field ($B$) normal to the plane of the surface.

Special Cases :

(i) When $\theta = 0^\circ$, i.e. uniform magnetic field is acting perpendicular to the plane of the surface, then

$\Phi_B = BA \cos 0^\circ$

 $(\because \cos 0^\circ = 1)$

$\Phi_B = BA = \text{maximum}$

Thus, magnetic flux through a given surface is maximum (i.e. maximum number of magnetic field lines pass through the given surface), when magnetic field is perpendicular to the plane of the surface.

(ii) When $\theta = 90^\circ$, i.e. uniform magnetic field is along the plane of the surface, then

$\Phi_B = BA \cos 90^\circ$

$(\because \cos 90^\circ = 0)$

$\Phi_B = 0$

Thus, the magnetic flux through a given surface is zero, when the magnetic field is along the plane of the surface.

Definition of Magnetic flux density or Strength of Magnetic field or Magnetic Induction (B) :

Since

$\Phi_B = BA$

$\therefore B = \frac{\Phi_B}{A}$

Thus, magnetic flux density (B) is defined as the magnetic flux (associated normally) per unit area.

Units of magnetic flux : 

In SI, magnetic flux is measured in weber (Wb)

Since

$\Phi_B = BA$

$1 \text{weber} = 1 Tesla \times metre^{2}$

$1 \text{Wb} = 1 T m^{2}$

In C.G.S. system, unit of magnetic flux is maxwell

$1 \text{ maxwell} = 1 \text{ G} \times 1 \text{ cm}^2$

$=10^{-4} \text{ T} \times 10^{-4} \text{ m}^2$

$= 10^{-8} \text{ T m}^2$

$= 10^{-8} \text{ weber}$

Dimensional formula for magnetic flux :

We know

$\Phi_B = BA \cos \theta$

Since

$B = \frac{F}{qv}$

and $\cos \theta$ is dimensionless

So,

$[\Phi_B] = [FA/qv]$

$(\because q = It)$

$[\Phi_B] = \left[ \frac{FA}{Ivt} \right]$

or

$[\Phi_B] = \frac{[F][A]}{[I][v]}$

$[\Phi_B] = \frac{[MLT^{-2}][L^2]}{[A][LT^{-1}]}$

$[\Phi_B] = [ML^2 T^{-2} A^{-1}]$

Nature of Magnetic Flux : Since $\Phi = \vec{B} \cdot \vec{A}$ i.e. dot product of $\vec{B}$ and $\vec{A}$, so $\Phi$ is a scalar physical quantity.

Magnetic Flux associated with variable magnetic field : 

We know, magnetic flux $\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta$ holds good only if the magnetic field is uniform over the surface. If the magnetic field ($\vec{B}$) is variable, then the surface is divided into number of area elements, each of which has area small enough to have constant magnetic field through it.

Let $\vec{B}_1, \vec{B}_2, \vec{B}_3, \dots \vec{B}_n$ be the magnetic field at the area elements of area $\Delta \vec{A}_1, \Delta \vec{A}_2, \Delta \vec{A}_3, \dots \Delta \vec{A}_n$ respectively of a surface (Figure 6), then the total magnetic flux through the surface is the sum of the magnetic flux through all area elements.

That is,

$\Phi_B = \vec{B}_1 \cdot \Delta \vec{A}_1 + \vec{B}_2 \cdot \Delta \vec{A}_2 + \dots + \vec{B}_n \cdot \Delta \vec{A}_n$

$\Phi_B= \sum \vec{B} \cdot \Delta \vec{A} \qquad ...(iv)$

If $\Delta A \rightarrow 0$, then eqn. (iv) can be written as

$\Phi_B = \int_S \vec{B} \cdot d\vec{A} \qquad ...(v)$

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