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Magnetic Flux | What Is Magnetic Flux

Magnetic Flux | What Is Magnetic Flux :- Magnetic flux represents the total number of magnetic field lines passing normally through a given surface. It depends on the strength of the magnetic field, the area of the surface, and its orientation with respect to the magnetic field.

A simple and intuitive way to visualize magnetic flux is by comparing it to rain falling through a rectangular loop. Imagine a uniform rain shower where the raindrops fall vertically downward, just like magnetic field lines passing through a surface. The rectangular loop represents the surface placed in this field.

When the loop is horizontal (θ = 0°), the rain falls directly through it. Maximum amount of rain passes through the loop per second — this represents maximum magnetic flux.
When the loop is tilted (at some angle θ), fewer raindrops pass through it because of the smaller effective area facing the rain. The flux decreases as the tilt increases.
When the loop is vertical (θ = 90°), the rain slides along the plane and none passes through — the magnetic flux becomes zero.

The amount of rain passing through the loop depends on the effective area of the loop as seen along the direction of the rain. From the figure, the effective area (i.e., the projection of the loop in the direction of rain) is

A_eff= ab cosθ = A cosθ

where A = ab is the actual area of the loop and is the tilt angle of the loop. A loop perpendicular to the rain (θ = 0^∘ has A_eff = A, meaning the full area is exposed. If the loop is tilted by , no raindrops pass through it, and A_eff = 0.

We can apply the same idea to a magnetic field passing through a loop. Below figure shows a loop of area A = ab placed in a uniform magnetic field.

Imagine the magnetic field lines (shown as dots (•) pointing out of the page) like arrows coming toward you. The density of these dots (number of dots per square meter) represents the strength of the magnetic field — a stronger field corresponds to dots packed more closely together.

The total number of dots (or field lines) passing through the loop depends on two factors :

The density of dots, which is proportional to , and
The effective area of the loop, A_eff= A cosθ .

The angle is the angle between the magnetic field (B) and the area vector (A) (axis of the loop). The maximum number of field lines passes through the loop when it is perpendicular to the magnetic field (). If the loop is tilted by 90°, no field lines pass through it.

Formula of Magnetic Flux

If θ is the angle between the direction of the field (B) and normal to the area, (area vector A) then, magnetic flus (φ) is defined as the dot product (scalar product) of magnetic filed (B) and the area vector (A).

$\displaystyle \phi =\overrightarrow{B}.\overrightarrow{A}=BAcos\theta$

If a coil has more than one turn, then the flux through the whole coil is the sum of the fluxes through the individual turns. If the magnetic field is uniform, then the total flux linkage,

$\displaystyle \phi =N\left( \overrightarrow{B}.\overrightarrow{A} \right)=NBAcos\theta$

Effect of Tilt Angle (θ) on Magnetic Flux

(1). When θ = 0°
The magnetic field is perpendicular to the loop (parallel to the area vector).

$\displaystyle \phi =BAcos0{}^\circ =0$

(2). When θ = 90°
The magnetic field is parallel to the plane of the loop (perpendicular to the area vector).

$\displaystyle \phi =BAcos90{}^\circ =0$

Unit and dimensions of Magnetic Flux

(1). SI unit of magnetic flux : weber (Wb).

(i) Using $\displaystyle \phi =BAcos\theta$ ,

1 weber = 1 Wb = T-m²

(ii) From equation F = BILsinθ (force on a current carrying concoctor placed in a magnetic field) , B = F/ILsinθ ⇒ 1 T = 1 NA^-1m^-1. Using this we get,

1 Wb = T-m²= (1 NA^-1m^-1) × m²= 1 NmA^-1= 1JA^-1

(iii) As V = W/Q ⇒ W = VQ ⇒ Joule = Volt × Coulomb,

1 Wb = 1JA^-1= 1 VCA^-1 = 1 V(As)A^-1

⇒ 1 Wb = 1 V-s

(iv) As V = IR ⇒ 1 Volt = 1 Amp. × Ohm,

1 Wb = 1 VCA^-1 = 1 (AΩ)CA^-1

⇒ 1 Wb = 1 ΩC

(2). CGS unit of magnetic flux : maxwell (Mx).

1 maxwell = 1 gauss-cm²

Relation between Maxwell and Weber

1 Wb = T-m²= 10⁴ gauss × 10⁴ cm²= 10⁸ gauss × cm²

⇒ 1 Wb = 10⁸Mx

(3). Dimensions of magnetic flux :

$\displaystyle [\phi ]=[BAcos\theta ]=[M{{T}^{-2}}{{A}^{-1}}][{{L}^{2}}]$

$\displaystyle \Rightarrow [\phi ]=[M{{L}^{2}}{{T}^{-2}}{{A}^{-1}}]$

✅ Note :

Magnetic field lines are imaginary—they are used only to represent the direction and strength of the magnetic field. However, magnetic flux is a real, measurable scalar physical quantity.

Example 1.

At certain location in the northern hemisphere, the earth’s magnetic field has a magnitude of 42 mT and points down ward at 57º to vertical. The flux through a horizontal surface of area 2.5 m² will be :

(A) 42 × 10^-6 Wb

(B) 22 × 10^-8 Wb

(D) 57 × 10^-6 Wb

(given cos 33º = 0.839, cos 57º = 0.545)

Solution :

$\displaystyle \phi =BAcos\theta =\left( 42\times {{10}^{-3}} \right)\times 2.5\times cos57{}^\circ$

$\displaystyle \therefore \phi =\left( 42\times {{10}^{-3}} \right)\times 2.5\times 0.545=57\times {{10}^{-6}}Wb$

Option (D) is correct.

Example 2.

A loop of wire is placed in a magnetic field $\displaystyle \overset{\to }{\mathop{B}}\,=0.3\,\hat{j}\,T$ . Find the flux through a loop of area $\displaystyle \overset{\to }{\mathop{A}}\,=(2\,\hat{i}+5\,\hat{j}\,-3\,\hat{k}){{m}^{2}}$ .

Solution :

$\displaystyle \phi =\overrightarrow{B}.\overrightarrow{A}=\left( 0.3\,\hat{j} \right).(2\,\hat{i}+5\,\hat{j}\,-3\,\hat{k})=1.5Wb$

Example 3.

A loop of area 0.06 m² is placed in a magnetic field 1.2 T with its plane inclined 30° to the field direction. Find the flux linked with plane of loop.

Solution :

Here θ = 90° – 30° = 60°

$\displaystyle \therefore \phi =BAcos60{}^\circ =1.2\times 0.06\times \frac{1}{2}=\text{0}\text{.036}Wb$

Example 4.

At a given plane, horizontal and vertical components of earth’s magnetic field B_H and B_V are along x and y axes respectively as shown in figure. What is the total flux of earth’s magnetic field associated with an area S, if the area S is in (a) x-y plane (b) y-z plane and (c) z-x plane ?

Solution :

(a) For area in x-y plane :

$\displaystyle \overrightarrow{S}=S\hat{k}$

$\displaystyle {{\phi }_{xy}}=(\hat{i}\,{{B}_{H}}-\hat{j}\,{{B}_{V}}).(\,\hat{k}S)=0$

(b) For area in y-z plane :

$\displaystyle \overset{\to }{\mathop{S}}\,=S\,\hat{i}$

$\displaystyle {{\phi }_{yz}}=(\hat{i}\,{{B}_{H}}-\hat{j}\,{{B}_{V}}).(\hat{i}\,S)={{B}_{H}}\,S$

(c) For area in z-x plane :

$\displaystyle \overset{\to }{\mathop{S}}\,=S\,\hat{j}$

$\displaystyle {{\phi }_{zx}}=(\hat{i}\,{{B}_{H}}-\hat{j}\,{{B}_{V}}).(\hat{j}\,S)=-{{B}_{V}}\,S$

Negative sign implies that flux is directed vertically downwards opposite to area vector S.

Magnetic Flux in a Non-uniform Magnetic Field

To find the magnetic flux through a non-uniform magnetic field, we consider a small area element of the loop. At each point, the magnetic field may have a different magnitude (and direction if the surface is non uniform). The small amount of magnetic flux through the element is :

$\displaystyle d\phi =\overrightarrow{B}.\overrightarrow{dA}=BdAcos\theta$

where is the angle between and the normal to the surface.

To obtain the total magnetic flux through the entire loop, we sum (integrate) the contributions from all such small elements :

$\displaystyle \phi =\int{d\phi =\int{BdAcos\theta }}$

Example 5.

A rectangular loop of dimensions 1.0 cm × 4.0 cm is placed 1.0 cm away from a long straight wire, as shown in the figure. The wire carries a current of 2.0 A. Find the magnetic flux passing through the loop.

Solution :

Magnetic field due to a long straight wire,

$\displaystyle B=\frac{{{\mu }_{o}}I}{2\pi r}$

Flux through a small strip of thickness dr of the loop,

$\displaystyle d\phi =BdA=\frac{{{\mu }_{o}}I}{2\pi r}\times \left( 0.04dr \right)$

Total magnetic flux through the loop,

$\displaystyle \phi =\int{d\phi =\int\limits_{0.01}^{0.02}{\frac{{{\mu }_{o}}I}{2\pi r}\times \left( 0.04dr \right)}}=\frac{0.02{{\mu }_{o}}I}{\pi }\int\limits_{0.01}^{0.02}{\frac{dr}{r}}$

$\displaystyle \Rightarrow \phi =\frac{0.02{{\mu }_{o}}I}{\pi }\left[ lo{{g}_{e}}(r) \right]_{0.01}^{0.02}=\frac{0.02{{\mu }_{o}}I}{\pi }lo{{g}_{e}}(\frac{0.02}{0.01})$

$\displaystyle \Rightarrow \phi =\frac{0.02{{\mu }_{o}}I}{\pi }lo{{g}_{e}}(2)=\frac{0.02\times \left( 4\pi \times {{10}^{-7}} \right)\times 2}{\pi }\times 2.303lo{{g}_{10}}(2)$