# Find the flux of Earth's magnetic field of magnitude 5.00 \times 10^{-5} T through a square loop...

## Question:

Find the flux of Earth's magnetic field of magnitude {eq}5.00 \times 10^{-5} {/eq} T through a square loop of area 20.0 cm{eq}^2 {/eq}

(a) when the field is perpendicular to the plane of the loop,

(b) when the field makes a 30.0{eq}^{\circ} {/eq} angle with the normal to the plane of the loop, and

(c) when the field makes a 90.0{eq}^{\circ} {/eq} angle with the normal to the plane.

## Magnetic Flux:

{eq}\\ {/eq}

When magnetic field lines pass through a surface, a magnetic flux is said to be associated with that surface, which is proportional to the number of field lines that pass through the surface along a net direction.

Mathematically, the magnetic flux through a plane surface placed in a uniform depends on the area of the plane surface, the component of the magnetic field perpendicular to the surface.

{eq}\\ {/eq}

We are given:

• The magnitude of the magnetic field, {eq}B=5.00\times 10^{-5}\;\rm T {/eq}
• The area of the loop, {eq}A=20.0\;\rm cm^2=2.00\times 10^{-3}\;\rm m^2 {/eq}

The magnetic flux through a loop of the area, {eq}A {/eq} placed in a uniform magnetic field of magnitude, {eq}B {/eq}, is given by the equation:

{eq}\phi=BA\cos\,\theta {/eq}

Here,

• {eq}\theta {/eq} is the angle between the magnetic field and the normal to the loop.

a)

If the field is perpendicular to the plane of the loop, then it must be parallel to the normal to the loop. Therefore, {eq}\theta=0^{\circ} {/eq}, and the magnetic flux becomes:

{eq}\begin{align*} \phi&=\left (5.00\times 10^{-5}\;\rm T \right )\left ( 2.00\times 10^{-3}\;\rm m^3 \right )\cos\,0^{\circ}\\ &=\boxed{1.00\times 10^{-7}\;\rm T.m^2} \end{align*} {/eq}

b)

If the angle between the normal and the magnetic field is {eq}\theta=30^{\circ} {/eq}, then the magnetic flux is:

{eq}\begin{align*} \phi&=\left (5.00\times 10^{-5}\;\rm T \right )\left ( 2.00\times 10^{-3}\;\rm m^3 \right )\cos\,30^{\circ}\\ &=\boxed{8.66\times 10^{-8}\;\rm T.m^2} \end{align*} {/eq}

c)

If the angle between the normal and the magnetic field is {eq}\theta=90^{\circ} {/eq}, then the magnetic flux is:

{eq}\begin{align*} \phi&=\left (5.00\times 10^{-5}\;\rm T \right )\left ( 2.00\times 10^{-3}\;\rm m^3 \right )\cos\,90^{\circ}\\ &=\boxed{0} \end{align*} {/eq} 