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3 years ago

11

To apply Problem-Solving Strategy 21.1 Faraday's law. A closely wound rectangular coil of 80 turns has dimensions 25.0 cm by 40.

0 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 T to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf E induced in the coil

2 answers:

Alisiya [41]3 years ago

8 0

Answer:

58.37 V

Explanation:

Given that

Number of winding on coil, N = 80 turns

Area of the coil, A = 25 cm * 40 cm = 0.25 m * 0.4 m

Magnitude of magnetic force, B = 1.1 T

Time of rotation, t = 0.06 s

See the attachment for calculations

wlad13 [49]3 years ago

8 0

Answer:

ℰ $Ф_{av}$ =58.37V

Explanation:

Given:

Number of winding on coil 'N' = 80 turns

Area 'A' = 25 cm x 40 cm = 0.25 m x 0.4 m =>0.10m²

Magnitude of magnetic force 'B' = 1.1 T

Time ' t' = 0.06 s

The average magnitude of the induced emf is given by:

ℰ $Ф_{av}$ =N|ΔФB/Δt| => N |Ф $Ф_{B,f}$ - Ф $Ф_{B,i}$ | /Δt

The flux through the coil is Ф $Ф_{B$ = BA cos∅

ℰ $Ф_{av}$ =NBA |cos(∅ $Ф_{f}$ ) - cos(∅ $Ф_{i}$ )| /Δt

As the initial angle is ∅= 97-37 =>53° and the final angle is ∅=0°

ℰ $Ф_{av}$ = $\frac{(80)(1.1)(0.1)|cos(0)-cos(53)|}{0.06}$

ℰ $Ф_{av}$ = $\frac{(8.8)|1-0.002|}{0.06}$

ℰ $Ф_{av}$ =58.37V

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