# A square loop of wire of side l = 5.0 cm is in a uniform magnetic field B = 0.16 T. What is the...

## Question:

A square loop of wire of side l = 5.0 cm is in a uniform magnetic field B = 0.16 T. What is the magnitude flux in the loop when B is an angle of 30 to the area of the loop?

## Magnetic Flux:

The magnetic field generates the magnetic flux in the conductor. The measurable unit used to express the magnetic flux is Weber. It is the surface integral of the magnetic lines.

Given data

• The value of the side length is {eq}l = 5\;{\rm{cm}} = {\rm{0}}{\rm{.05}}\;{\rm{m}} {/eq}
• The value of the magnetic field is {eq}B = 0.16\;{\rm{T}} {/eq}
• The value of the angle is {eq}\theta = 30^\circ {/eq}

The expression for the flux is,

{eq}\begin{align*} {\phi _{flux}} &= BA\sin \theta \\ {\phi _{flux}} &= B\left( {{l^2}} \right)\sin \theta \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {\phi _{flux}} &= \left( {0.16} \right)\left( {{{0.05}^2}} \right)\sin 30^\circ \\ &= 2 \times {10^{ - 4}}\;{\rm{Weber}} \end{align*} {/eq}

Thus, the value of the magnetic flux is {eq}2 \times {10^{ - 4}}\;{\rm{Weber}} {/eq}