# A five-turn circular wire coil of radius 0.425 m lies in a plane perpendicular to a uniform...

## Question:

A five-turn circular wire coil of radius 0.425 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.370 T. If the wire is reshaped from a five-turn circle to a three-turn circle in 0.101 s (while remaining in the same plane), what is the average induced emf in the wire during this time?

## Magnetic flux:

In physics, the term magnetic flux is the Number of magnetic lines in a close body. It measures the magnetic field in a closed surface area of an object like a solenoid. The magnetic flux of the closed surface is always zero.

Given data

• Radius of the coil {eq}r = 0.425\;{\rm{m}}.{/eq}
• The magnitude of magnetic field {eq}B = 0.370\;{\rm{T}}.{/eq}
• Time taken to reshaped {eq}t = 0.101\;{\rm{s}} .{/eq}

The expression for the flux is

{eq}\phi = NAB\cos \theta {/eq}

Substituting the given values,

Initial flux is

{eq}\begin{align*} {\phi _i} &= 5\left( {\pi \times {{\left( {0.425} \right)}^2}} \right) \times 0.370\\ {\phi _i} &= 1.049\;{\rm{wb}} \end{align*} {/eq}

Since the coil is reshaped, the initial length is equal to the final length.

{eq}3 \times \left( {2\pi {r_n}} \right) = 5 \times \left( {2\pi r} \right) {/eq}

Substituting the given values,

{eq}\begin{align*} 3 \times \left( {2\pi {r_n}} \right) &= 5 \times \left( {2\pi \left( {0.425} \right)} \right)\\ {r_n} &= 0.708\;{\rm{m}} \end{align*} {/eq}

Therefore, final flux is

{eq}{\phi _f} = 3 \times \left( {\pi {r_n}^2} \right) \times B {/eq}

Substituting the given values,

{eq}\begin{align*} {\phi _f} &= 3\pi {\left( {0.708} \right)^2} \times \left( {0.370} \right)\\ {\phi _f} &= 1.74\;{\rm{wb}} \end{align*} {/eq}

The expression for the induced EMF is,

{eq}E = - \dfrac{{\Delta \phi }}{{\Delta t}} {/eq}

Substituting the given values,

{eq}\begin{align*} E &= - \dfrac{{\left( {1.74 - 1.049} \right)}}{{\left( {0.101} \right)}}\\ E &= - 6.841\;{\rm{V}} \end{align*} {/eq}

Therefore, the induced EMF is -6.841 V.