Physics, 23.07.2019 13:30 aidananderson

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 =

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Answer from: jholland03

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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Answer from: elizabethhubbe

(a) The magnetic flux through the coil is given by
$\Phi = N B A \cos \theta$
where
N is the number of turns in the coil
B is the intensity of the Earth's magnetic field
A is the area enclosed by one coil
$\theta$ is the angle between the direction of B and the perpendicular to A.

Initially, the coil is perpendicular to the Earth's magnetic field, therefore $\theta=0^{\circ}$ and $\cos \theta=1$ , so the magnetic flux through the coil is given by:
$\Phi_i = NBA = (200) (5.20 \cdot 10^{-5} T)(10.8 \cdot 10^{-4} m^2)=1.12 \cdot 10^{-5}Wb$

(b) After the coil is rotated, its plane is parallel to the magnetic field. This means that $\theta=90^{\circ}$ and $\cos \theta=0$ , so the magnetic flux through the coil is simply zero:
$\Phi_f = NBA (0) = 0$

(c) The variation of magnetic flux through the coil is
$\Delta \Phi = \Phi_f - \Phi_i = 0-(1.12 \cdot 10^{-5} Wb) = -1.12 \cdot 10^{-5} Wb$
we can ignore the negative sign since we are interested only in the magnitude of the average emf, not in its direction.

For Faraday-Neumann-Lenz law, the average emf induced in the coil is given by
$\epsilon = \frac{\Delta \Phi}{\Delta t}= \frac{1.12 \cdot 10^{-5} Wb}{5.00 \cdot 10^{-2}s}=2.25 \cdot 10^{-4} V$

Answer from: Quest

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Answer from: Quest

connect the batteries in series and hope

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