Question

A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a position in which it makes an angle of 37.0 degrees with a magnetic field of 1.10 \rm T to a position perpendicular to the field. The rotation takes 0.0600 \rm s. What is the average emf EMF induced in the coil?What is the magnitude of the average emf E induced as the coil is rotated?E= ________ V

Answer 1

Answer:

88.3

Explanation:

Emf in a rotating coil is given by rate of change of flux:

E= dФ/dt=(NABcos∅)/ dt

N: number of turns in the coil= 80

A: area of the coil= 0.25×0.40= 0.1

B: magnetic field strength= 1.1

Ф: angle of rotation= 90- 37= 53

dt= 0.06s

E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V

Answer 2

Answer:

E =90.25V

Explanation:

The illustration and formula to use in in the attachment.

• The Turn is N=80

• The dimension is 25cm by 40cm,  A=0.10$m^{2}$

• Magnetic field of B is 1.7 T

• Angle is ∅=80°

• and time is Δt=0.0600 s

$E_{av} =$ $\frac{N*B*A IcosTita_{f}-cosTita_{i} }{Dt}$

      =$\frac{80*1.70*0.10(cos(0)-cos(53)}{0.0600}$

     =$\frac{5.415}{0.0600}$

$E_{av}$ = 90.25V