Answer:
induced.
Step-by-step explanation:
We define all our variables,
![B=1.2T](https://img.qammunity.org/2020/formulas/physics/college/iux6conp1c3nk5lrssfzof7h6dyt4hi5u5.png)
![A=0.27m*0.50m](https://img.qammunity.org/2020/formulas/physics/college/y2pct9u8exxdqh39bvf0qytjwru9tt4rfv.png)
![\theta= 90-37](https://img.qammunity.org/2020/formulas/physics/college/86o2n9199z34yj5s8zhpw0ys450tchq46o.png)
![\Delta T= 0.09s](https://img.qammunity.org/2020/formulas/physics/college/tdeqftdkggqkviw417dp562jixlr4cexgc.png)
![N= 100](https://img.qammunity.org/2020/formulas/physics/college/kakhdpvvmxi1zoiomepr4vbcvtmdli8z2t.png)
EMF induced is given through the expression
![\epsilon = (-N\Delta \Phi)/(\Delta t)](https://img.qammunity.org/2020/formulas/physics/college/noi3fzvzsbsbsfj4c6tgjp2o734ig57g9k.png)
Here we understand
as
![\Phi = BAcos\theta](https://img.qammunity.org/2020/formulas/physics/college/gyxhaftlhu7rxou81llm9pwmf43tn282g4.png)
We proceed to calculate the entire Initial Flow as the final, as well
![\Phi_i=(1.2)(0.27*0.5)cos(90-37) = 0.09749Tm^2](https://img.qammunity.org/2020/formulas/physics/college/vi6341cdxkr4b8ccljo7dho6s603eylhie.png)
Final Flow
![\Phi_f=(1.2)(0.27*0.5)= 0.162Tm^2](https://img.qammunity.org/2020/formulas/physics/college/u81xsuc1hwy80cry20lae95sun2hassbco.png)
Now, if
,
![\epsilon= (100)(0.162Tm^2 - 0.09749Tm^2)/(0.09s)](https://img.qammunity.org/2020/formulas/physics/college/m4dhyac9xu9aox9qlqh1xcoekgtx6bux4u.png)
induced.
NOTES:
- It is necessary to make two small notes regarding the development of the exercise. The subtraction of the angles is used since the exercise indicates that the angle is between the field B and the Plane. However, the measurement between the Area and the field is required b.
- Negative signs can be neglected because it is understood that this is a reference to know which direction has the highest potential.