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liubo4ka [24]
2 years ago
12

Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the bea

m, and moves at 314 m/s. What is the maximum magnetic force it can feel
Physics
1 answer:
mylen [45]2 years ago
5 0

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2

The maximum magnetic field is calculated using the following intensity formula;

I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

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I need help in physics please help me
artcher [175]

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For the vector C we have a magnitude of 4.8 and an angle 22 ° with the axis -y (direction j)

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x = 4.8sin(22)i\\\\y = 4.8cos(22)(-j)

So:

C = 4.8sin(22) i + 4.8cos(22)(-j)\\\\C = 1.798\ i - 4.450\ j

For Vector B we have a magnitude of 5.6 and an angle of 33 with the -x axis (-i direction)

So:

x = 5.6cos(33)(-i)\\\\y = 5.6 sin(33)(-j)

So:

B = 5.6cos(33)(-i) + 5.6sin(33)(-j)\\\\B = -4.696\ i - 3.05\ j

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F = (-4.696 +1.798)i + (-4.450 - 3.05)j\\\\F = -2.898\ i - 7.5\ j

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3 years ago
You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen
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Answer:

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\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s  \right )D+t_t^2v_s^2=0

<em>(B)  D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

\displaystyle D=\frac{gt^2}{2}

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\displaystyle t=\sqrt{\frac{2D}{g}}

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

\displaystyle t=\frac{D}{v_s}

(A)

The total measured time is the sum of both times and it's given as t_t=3.5\ seconds

\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}

From this equation we'll manage to compute D

First, we isolate the square root

\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}

Let's square both sides

\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}

Multiplying by v_s^2

\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2

Rearranging and factoring

\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}

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Computing all the coefficients:

\displaystyle D^2-26,705.82D+1,458,056.25=0

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D=54.71,\ D=26,651.11

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<em>D=54.71 m</em>

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