In a physics laboratory experiment, a coil with 200 turns enclosing an area of 10.8 cm2 is rotated during the time interval 5.00×10−2 s from a position in which its plane is perpendicular to earth's magnetic field to one in which its plane is parallel to the field. the magnitude of earth's magnetic field at the lab location is 5.20×10−5 t . part a what is the magnitude of the magnetic flux (φinitial) through the coil before it is rotated? express your answer numerically, in webers, to at least three significant figures. view available hint(s) |φinitial| = wb submit part b what is the magnitude of the magnetic flux φfinal through the coil after it is rotated? express your answer numerically, in webers, to at least three significant figures. view available hint(s) |φfinal| = wb submit part c what is the magnitude of the average emf induced in the coil? express your answer numerically, in volts, to at least three significant figures. view available hint(s) average induced emf =

A) The magnitude of the magnetic flux (φ_initial) through the coil before it is rotated is; φ_i = 1.123 × 10⁻⁵ Wb

B) The magnitude of the magnetic flux (φ_final) through the coil after it is rotated is; φ_F = 0 Wb

C) The magnitude of the average emf induced in the coil is;

ε = 22.5 × 10⁻⁵ Wb

We are given;

Number of turns; N = 200 turns

Area; A = 10.8 cm² = 0.00108 m²

Magnitude of magnetic field; B = 5.2 × 10⁻⁵ T

Time; t = 5 × 10⁻² s = 0.05 s

A) Formula for the magnitude of the magnetic flux (φ_initial) through the coil before it is rotated is;

φ_i = NAB

Plugging in the relevant values gives;

φ_i = 200 × 0.00108 × 5.2 × 10⁻⁵

φ_i = 1.123 × 10⁻⁵ Wb

B) Formula for the magnitude of the magnetic flux (φ_final) through the coil after it is rotated is;

φ_f = NAB cos θ

After it is rotated, θ = 90°

Thus;

φ_f = 200 × 0.00108 × 5.2 × 10⁻⁵ × cos 90°

φ_f = 0 Wb

C) The magnitude of the average emf induced in the coil is;

ε = -(φ_f - φ_i)/t

Thus;

ε = -(0 - 1.123 × 10⁻⁵)/0.05

ε = 22.46 × 10⁻⁵ Wb

To 3 significant figures;

ε = 22.5 × 10⁻⁵ Wb

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i believe the answer is a

wavelength divides in half and speed stays the same.