Magnetic Flux

Electromagnetic Induction · CBSE

Theory

The magnetic flux through a plane surface of area $A$ placed in a uniform magnetic field $\vec{B}$ is the scalar product of the field and the area vector.

$\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$

$\theta$ is the angle between $\vec{B}$ and the AREA VECTOR $\vec{A}$ , which is drawn along the normal to the surface, with magnitude equal to the area.
Equivalently, flux is the total number of magnetic field lines crossing the surface normally.
Magnetic flux is a SCALAR quantity. SI unit: weber (Wb), where $1\ \text{Wb} = 1\ \text{T m}^2$ . Dimensions $[\text{M L}^2\text{T}^{-2}\text{A}^{-1}]$ .

If $\vec{B}$ has different magnitudes and directions over different parts of the surface, divide the surface into small elements $d\vec{A}_i$ , each small enough that $\vec{B}_i$ is uniform over it, and add up.

$\Phi_B = \vec{B}_1\cdot d\vec{A}_1 + \vec{B}_2\cdot d\vec{A}_2 + \cdots = \sum_{\text{all}} \vec{B}_i\cdot d\vec{A}_i$

What the question states	Angle to use in $\cos\theta$	Flux
Coil placed perpendicular to $\vec{B}$ (plane normal to $B$ )	$\theta = 0^{\circ}$	$\Phi_B = BA$ , maximum
Plane of the coil contains $\vec{B}$	$\theta = 90^{\circ}$	$\Phi_B = 0$
PLANE of the coil makes angle $\alpha$ with $\vec{B}$	$\theta = 90^{\circ}-\alpha$	$\Phi_B = BA\sin\alpha$
NORMAL of the coil makes angle $\theta$ with $\vec{B}$	$\theta$ as given	$\Phi_B = BA\cos\theta$

Worked instance (CBSE 2020-21): a coil of area $100\ \text{cm}^2$ has its PLANE at $60^{\circ}$ to a field of $10^{-1}\ \text{T}$ . Then the normal is at $\theta = 90^{\circ}-60^{\circ}=30^{\circ}$ , so

$\Phi_B = BA\cos 30^{\circ} = 10^{-1}\times 100\times 10^{-4}\times \dfrac{\sqrt{3}}{2} \approx 0.866\times 10^{-3}\ \text{Wb}$

If that field then falls to zero in $10^{-3}\ \text{s}$ , the induced emf is $\varepsilon = A\cos\theta\dfrac{dB}{dt} = 100\times10^{-4}\times\cos 30^{\circ}\times 10^{2} = 0.866\ \text{V}$ .

For a closely wound coil of $N$ turns the same flux $\Phi_B$ threads every turn, so the total flux linkage is $N\Phi_B$ , measured in weber-turns.
Flux linkage, not flux, is what appears in $N\Phi_B = LI$ and $N_1\Phi_1 = MI_2$ , and in $\varepsilon = -N\,d\Phi_B/dt$ .

Question (Delhi 2017): a long straight current-carrying wire passes NORMALLY through the centre of a circular loop. If the current in the wire increases, is an emf induced in the loop?
Answer: the field lines of a straight wire are concentric circles lying in planes perpendicular to the wire. Since the wire is normal to the loop, the plane of the loop is itself one such plane, so at every point of the loop $\vec{B}$ is tangential, i.e. it lies IN the plane of the loop.
Hence the angle between $\vec{B}$ and the loop's area vector (which points along the wire) is $90^{\circ}$ , and

$\Phi_B = BA\cos 90^{\circ} = 0 \quad \text{at all times}$

The flux is zero however large the current becomes, so there is no CHANGE of flux and therefore NO induced emf in the loop. The answer turns entirely on geometry, not on the size of the current.

$\Phi_B = BA\cos\theta$ ; maximum $BA$ , minimum $0$ ; scalar; unit Wb.
Rectangular loop of width $l$ with length $x$ inside a field region: $\Phi_B = Blx$ .
Circular coil of radius $r$ , $N$ turns, field along the axis: $\Phi_B = N B \pi r^2$ .
Solenoid of $N$ turns, length $l$ , area $A$ , current $i$ : $\Phi_{\text{linkage}} = NBA = \dfrac{\mu_0N^2iA}{l}$ .

Real board questions. Answers are at the end of the page.

1
A long straight current carrying wire passes normally through the centre of a circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify.
1 markCBSE Class 12 Physics — Very Short Answer · Delhi 2017 · Section A, 1 markAnswer 1 ↓

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
The magnetic field lines due to the straight current-carrying wire are concentric circles lying in planes perpendicular to the wire. Since the wire passes normally (perpendicular) through the centre of the loop, the plane of the loop is itself one such plane perpendicular to the wire, so at every point of the loop the field $\vec{B}$ is tangential, i.e. it lies in the plane of the loop. Hence the angle between $\vec{B}$ and the area vector $\vec{A}$ of the loop (which is normal to the loop's plane, i.e. along the wire) is $90°$ , so the flux linked with the loop is $\phi = BA\cos 90° = 0$ at all times. Since the flux remains zero even as the current (and hence $B$ ) increases, there is no change in flux, and therefore no induced emf is set up in the loop.