# A generator has a square coil consisting of 202 turns. The coil rotates at 75.7 rad/s in a...

## Question:

A generator has a square coil consisting of 202 turns. The coil rotates at 75.7 rad/s in a 0.219-T magnetic field. The peak output of the generator is 71.9 V. What is the length of one side of the coil?

## Lenz's law:

Lens introduced the direction of the induced current due to the changing flux in the circuit. Lenz's law states that the induced current is directed is such direction which is opposing the change in the flux throughout the circuit means if a time changing current is producing a flux then the induced current will flow in opposite direction of that time changing current.

Given:

• The number of turns {eq}N=202 \ \text{turns} {/eq}
• The angular velocity {eq}\omega=75.7 \ \text{rad/s} {/eq}
• The magnetic field {eq}B_0=0.219 \ \text{T} {/eq}
• The peak output voltage (induced emf) {eq}\varepsilon=71.9 \ \text{V} {/eq}

Suppose the time-dependent magnetic field is {eq}B=B_0\text{sin}(\omega t) {/eq}

From the Lenz's law, the induced emf is formulated as:

{eq}\begin{align} \left|\varepsilon\right|&=\left|N\frac{\mathrm{d} \phi}{\mathrm{d} t}\right|\\ \\ &=N\frac{\mathrm{d} }{\mathrm{d} t}(BA)\\ \\ &=NA\frac{\mathrm{d} }{\mathrm{d} t}(B_0\text{sin}(\omega t))\\ \\ &=NAB_0\omega\text{cos}(\omega t))\\ \end{align} {/eq}

The peak output emf is:

{eq}\begin{align} \varepsilon=N(L^2)B_0\omega\\ \end{align} {/eq}

Thus the length of the one side of the coil is:

{eq}\begin{align} L&=\sqrt{\frac{\varepsilon}{NB_0\omega}}\\ &=\sqrt{\frac{(71.9 \ \text{V})}{(202 \ \text{turns})(0.219 \ \text{T})(75.7 \ \text{rad/s})}}\\ &=\color{blue}{0.146 \ \text{m}} \end{align} {/eq}