### What is Magnetic Flux:

Magnetic flux is defined as the number of magnetic field lines passing through a given closed surface. It provides a measure of the total magnetic field passing through a given area. Here, the region under consideration can be of any size and in any orientation with respect to the direction of the magnetic field.

Magnetic flux is defined as the total number of magnetic field lines through a given coil or section. It is the common component of the magnetic field passing through the coil. The magnetic flux is denoted by ΦB, where B is a magnetic field and its unit is Weber (Wb).

The magnetic flux value depends on the direction of the magnetic field and is a vector quantity.

### Magnetic Flux Formula:

Magnetic Flux formula is Φ_{B(wb)} in weber is equal to the magnetic field B_{(T)} in Tesla and multiply the area A_{(m2)} in meter^{2} and multiply the angle at which the field lines pass through the given surface area cos θ_{(d)} in degree.Hence the magnetic flux formula can be written as

Φ_{B(wb)}= B_{(T)} * A_{(m2)} * cos θ_{(d)}

### Derivation:

Φ_{B}= B * A

Φ_{B}= B * A cos θ

Φ_{B} = is the magnetic Flux in Weber(Wb)

B= is the magnetic field inTelsa

A = is the area in meter^{2}

Θ = the angle at which the field lines pass through the given surface area in degree

### Example:1

The Dimension of a rectangular loop is 0.80 m and 0.40cm.B and Θ are 009T and 65^{0} respectively.Determine the magnetic flux through the surface?

### Answer:

Given

Dimensions of rectangular loop =0.80 m and 0.40m

B =0.09T

Θ =65^{0}

Magnetic flux formula is given by

Φ_{B}= B * A cos θ

Area =0.80* 0.40

=0.32 m^{2}

Φ_{B}=0.09*0.32 * cos 65^{0}

Φ_{B}=0.0155Wb

### Example:2

The Dimension of a rectangular loop is 2.30 m and 2.10cm.B and Θ are 245T and 30^{0} respectively.Determine the magnetic flux through the surface?

### Answer:

Given

Dimensions of rectangular loop =0.80 m and 0.40m

B =245T

Θ =30^{0}

Magnetic flux formula is given by

Φ_{B}= B * A cos θ

Area =2.30* 2.10

=4.83 m^{2}

Φ_{B}=245*4.83 * cos 30^{0}

Φ_{B}=639.36Wb