# A 100-turn square wire coil of area 0.040 m2 (20.0 cm x 20.0 cm) rotates about a vertical axis at...

## Question:

A 100-turn square wire coil of area 0.040 m{eq}^2 {/eq} (20.0 cm {eq}\times {/eq} 20.0 cm) rotates about a vertical axis at 1500 rev/min. The horizontal component of Earth's magnetic field at the location of the loop is {eq}2.0 \times 10^{-5} {/eq} T. Calculate the maximum emf induced in the coil by Earth s field.

## Magnetic Field:

The magnetic field is a physical quantity that can be defined as a region of space in which if a particle with charge is moving then the charge feels the magnetic force on it. The basic standard measurement unit of the magnetic field is Tesla.

Given data:

• The magnetic field is {eq}B = 2.0 \times {10^{ - 5}}\,{\rm{T}} {/eq}
• The area is {eq}a = 0.040\,{{\rm{m}}^2} {/eq}
• The number of turns is {eq}N = 100 {/eq}
• The angular speed is {eq}w = 1500\,{\rm{rev/min}} = \left( {1500\,{\rm{rev/min}}} \right) \times \left( {\dfrac{{2\pi }}{{60}}} \right)\,{\rm{rad/s}} = 157.08\,{\rm{rad/s}} {/eq}

The expression for the maximum Emf induced in the coil is given by

{eq}e = NBaw {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} e &= NBaw\\ e &= \left( {100} \right)\left( {2.0 \times {{10}^{ - 5}}} \right)\left( {0.040} \right)\left( {157.08} \right)\\ e &= 0.0126\,{\rm{V}} \end{align*} {/eq}

Thus the emf induced in the coil is {eq}e \approx 0.013\,{\rm{V}} {/eq}