Question

A closely wound rectangular coil of 100 turns has dimensions of 27.0 cm by 50.0 cm . The plane of the coil is rotated from a position where it makes an angle of 37.0 ∘ with a magnetic field of 1.20 T to a position perpendicular to the field. The rotation takes 9.00×10−2 s . (Note that the coil is perpendicular to the area vector of the coil.) What is the average EMF induced in the coil?

Answer 1

Answer:

$\epsilon =71.677V$ induced.

Explanation:

We define all our variables,

$B=1.2T$

$A=0.27m*0.50m$

$\theta= 90-37$

$\Delta T= 0.09s$

$N= 100$

EMF induced is given through the expression

$\epsilon = \frac{-N\Delta \Phi}{\Delta t}$

Here we understand $\Phi$ as

$\Phi = BAcos\theta$

We proceed to calculate the entire Initial Flow as the final, as well

$\Phi_i=(1.2)(0.27*0.5)cos(90-37) = 0.09749Tm^2$

Final Flow

$\Phi_f=(1.2)(0.27*0.5)= 0.162Tm^2$

Now, if $\Delta t = 0.09s$,

$\epsilon= (100)(0.162Tm^2 - 0.09749Tm^2)/(0.09s)$

$\epsilon =71.677V$ 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.