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3 years ago

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A laser emits light at power 6.20 mW and wavelength 633 nm. The laser beam is focused (narrowed) until its diameter matches the

1080 nm diameter of a sphere placed in its path. The sphere is perfectly absorbing and has density 5.00 × 103 kg/m3. What are (a) the beam intensity at the sphere's location, (b) the radiation pressure on the sphere, (c) the magnitude of the corresponding force, and (d) the magnitude of the acceleration that force alone would give the sphere

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa , c) F = 20.6 10⁻¹² N

Explanation:

a) The intensity defined as the energy per unit area

S = U / A

Area of a circle is

W = 6.2 mw = 6.2 10-3 W

R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m

A = π R2

A = π (1,080 10⁻⁶)²

A = 3.66 10 -12 m²

S = 6.2 10-3 / 3.66 10-12

S = 1.69 10⁹ W / m²

b) The radiation pressure

P = 1 / c (dU / dt) / A

S = (dU / dt) / A

P = S / c

P = 1.69 10 9 / 3. 108

P = 5.63 Pa

c) the definition of pressure is force over area

P = F / A

F = P A

F = 5.63 3.66 10⁻¹²

F = 20.6 10⁻¹² N

d) for this we use Newton's second law

F = ma

a = F / m

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Explanation:

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By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

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However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

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Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

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$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

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