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Softa [21]
3 years ago
14

What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Physics
1 answer:
Aleksandr-060686 [28]3 years ago
5 0

Complete Question

The maximum electric field strength in air is 4.0 MV/m. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Answer:

The  value  is  P = 3.270960 *10^{6} \  W

Explanation:

From the question we are told that

   The  electric field strength is  E  = 4.0 \  M \  V/m  =  4.0 *10^6 \  V/m

    The  diameter is  d =  1.4 \  cm  = 0.014 \ m

Generally the radius is mathematically represented as

        r = \frac{d}{2}

=>     r = \frac{0.014}{2}

=>    r   = 0.007 \  m

Generally the cross-sectional area is mathematically represented as

        A =  \pi r^2

        A =  3.142 *  (0.007)^2

       A = 0.000154 \ m^2

Generally the maximum power that can be delivered is mathematically represented as

        P = \frac{c *  \epsilon_o  *  E^2 *  A }{2}

Here c is the speed of light with value  c =  3.0*10^{8} \  m/s

        \epsilon_o is the permittivity of free space with value  \epsilon_o  =  8.85 *10^{-12}  \ m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

     P =  \frac{3,0*10^8 *  8.85*10^{-12} *  (4 *10^6)^2 * 0.00154}{ 2}

       P = 3.270960 *10^{6} \  W

         

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Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
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Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
3 years ago
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