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Agata [3.3K]
3 years ago
15

Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of

the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?
Physics
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r    where i₁ and i₂ are current in the two wires , r is distance between the two and  μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i²  = 3.98 x 10⁻²

i = 1.995 x 10⁻¹

= .1995

=  0.2 A approx .

2 i = .4 A Ans .

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3 years ago
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Laskar, J.: 1990, The chaotic motion of the solar system. A numerical estimate of the size of the chaotic zones, Icarus, 88, 266
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1 year ago
Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
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Answer:

145 m

Explanation:

Given:

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we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

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Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

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Label A:

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