Answer:
The maximum emf that can be generated around the perimeter of a cell in this field is
![1.5732*10^(-11)V](https://img.qammunity.org/2020/formulas/physics/college/b50auxqu0eg7h82k9bydjq2waivmteed6w.png)
Step-by-step explanation:
To solve this problem it is necessary to apply the concepts on maximum electromotive force.
For definition we know that
![\epsilon_(max) = NBA\omega](https://img.qammunity.org/2020/formulas/physics/college/gcnj06i2o1l735zyckmzs83jcxrmbjjmjj.png)
Where,
N= Number of turns of the coil
B = Magnetic field
Angular velocity
A = Cross-sectional Area
Angular velocity according kinematics equations is:
![\omega = 2\pi f](https://img.qammunity.org/2020/formulas/physics/middle-school/pwzsovla2h22uvtixjbu2t72piygxnka5a.png)
![\omega = 2\pi*61.5](https://img.qammunity.org/2020/formulas/physics/college/3wvq9yi9qzhfboeci71s0ycql7j98cqm7j.png)
![\omega =123\pi rad/s](https://img.qammunity.org/2020/formulas/physics/college/tnwtsz2fswwqvos9ba38xu5dg3b5e6nwy9.png)
Replacing at the equation our values given we have that
![\epsilon_(max) = NBA\omega](https://img.qammunity.org/2020/formulas/physics/college/gcnj06i2o1l735zyckmzs83jcxrmbjjmjj.png)
![\epsilon_(max) = NB(\pi ((d)/(2))^2)\omega](https://img.qammunity.org/2020/formulas/physics/college/21ng5ot5glg3p4dh5l2dwgjlwzyp6quxa6.png)
![\epsilon_(max) = (1)(1*10^(-3))(\pi ((7.2*10^(-6))/(2))^2)(123\pi)](https://img.qammunity.org/2020/formulas/physics/college/28rt3wqxc3rzmynm50h5ye61dr61ioqmpi.png)
![\epsilon_(max) = 1.5732*10^(-11)V](https://img.qammunity.org/2020/formulas/physics/college/3xvbov9rs3bcfc73y3g0i8bu6njohgch9d.png)
Therefore the maximum emf that can be generated around the perimeter of a cell in this field is
![1.5732*10^(-11)V](https://img.qammunity.org/2020/formulas/physics/college/b50auxqu0eg7h82k9bydjq2waivmteed6w.png)