# The top part of the wire is bent into a semicircle of radius r = 0.29 m. The normal to the plane...

## Question:

The top part of the wire is bent into a semicircle of radius {eq}r = 0.29 m {/eq}. The normal to the plane of the loop is parallel to a constant magnetic field {eq}(\varphi = 0^{\circ}) {/eq} of magnitude {eq}0.75 T {/eq}. What is the change {eq}\Delta \phi {/eq} in the magnetic flux that passes through the loop when, starting with the position, the semicircle is rotated through half a revolution?

## Magnetic flux:

The term magnetic flux can be defined as the dot product of the magnetic field and the area of the given object. It's generally measured in Weber. The magnetic field is a scalar quantity.

Given data:

• The radius of the semi-circle is {eq}r = 0.29\,{\rm{m}} {/eq}
• The magnitude of the magnetic field is {eq}B = 0.75\,{\rm{T}} {/eq}

The expression for the change in the magnetic flux is given by

{eq}\Delta \phi = B \cdot \left( {{a_2} - {a_1}} \right) {/eq}

• Here {eq}{a_2} = \pi {r^2} {/eq} is the area of the semi-circle.
• Here {eq}{a_1} = 0 {/eq} is the area of the semi circle as starting with the position.

Substituting the values in the above equation as,

{eq}\begin{align*} \Delta \phi &= B \cdot \left( {{a_2} - {a_1}} \right)\\ \Delta \phi &= 0.75 \cdot \left( {\pi {r^2}} \right)\\ \Delta \phi &= 0.75 \times \left( {3.14 \times {{\left( {0.29} \right)}^2}} \right)\\ \Delta \phi &= 0.198055\,{\rm{T}}{{\rm{m}}^2} \end{align*} {/eq}

Thus, the change in the magnetic flux is {eq}\Delta \phi = 0.198055\,{\rm{T}}{{\rm{m}}^2} {/eq} 