Question

A closely wound rectangular coil of 80 turns has dimensions of 25.0 cm by 40.0 cm. The plane of the coil is rotated from a position where it makes an angle of 41.0 ∘ with a magnetic field of 1.50 T to a position perpendicular to the field. The rotation takes 0.0800 s .Part AWhat is the average emf induced in the coil?Express your answer with the appropriate units.

Answer 1

Answer:

$\varepsilon_{prom}=51.59V$

Explanation:

1) Notation and data given

N= 80 represent the turns

B=1.5 T represent the magnetic field

Dimensions =25cm x40cm

$\phi=41$° represent the angle respect to the plpane of the coil

$\phi_i=90-41=49$° since we need the angle respect to the magnetic field

$\phi_f =0$° since the final position is perpendicular to the field.

$\Delta t= 0.08s$

$\Phi_{B,f}$ represent the final flux through the coil

$\Phi_{B,i}$ represent the initial flux through the coil

$\varepsilon$ represent the induced emf, known as "electromagnetic induction" and is defined as "the production of voltage in a coil because of the change in a magnetic flux through a coil" (Variable of interest).

2) Formulas to use

We can begin calculating the area given by:

$A=0.25mx0.40m=0.1m^2$

We can use the formula for the average magnitude when we have an induced emf, given by:

$\varepsilon_{prom}=N|\frac{N\Phi_B}{\delta t}|=N|\frac{\Phi_{B,f}-\Phi_{B,i}}{\delta t}|$   (1)

We have another formula for the flux through the coil given by:

$\Phi_B =BAcos(\phi)$

Replacing this into equation (1) we got:

$\varepsilon_{prom}=\frac{NBA|cos(\phi_f)-cos(\phi_i)|}{\Delta t}$   (2)

3) Calculate the final answer

Now we can replace all the values given into equation (2) like this:

$\varepsilon_{prom}=\frac{(80)(1.5T)(0.1m^2)|cos(0)-cos(49)|}{0.0800s}=51.59V$