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velikii [3]
3 years ago
9

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

Physics
2 answers:
Law Incorporation [45]3 years ago
5 0

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=0.10m^{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

Pepsi [2]3 years ago
4 0

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

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