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3 years ago

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The current in a long solenoid of radius 4 cm and 19 turns/cm is varied with time at a rate of 8 A/s. A circular loop of wire of

radius 6 cm and resistance 3 Ω surrounds the solenoid. Find the electrical current induced in the loop (in µA).

1 answer:

Dmitry [639]3 years ago

4 0

Answer:

Current induced in the loop = 0.032 mA

Explanation:

emf induced in the solenoid using Faraday law e = $\frac{-d\phi}{dt}=-\pi R^{2} \frac{dB}{dt}$

here, R is the radius of solenoid which is constant & $\frac{dB}{dt}$ is the change in magnetic field.

Magnetic field inside the solenoid from B=μ0 n i -----(i)

Where, i is the current in the coil.

n= numbers of turn in per meter length.

n=19 x 100 turns/m

Differentiating equation (i)

$\frac{dB}{dt} =$ μ0 n $\frac{dI}{dt}$

= 4 π 10-7 x 1900 x 8 =0.019

$E = 3.14 X 0.42^{2} X 0.019$ =0.095 m V

Hence electrical current induced in the loop = $\frac{E}{R} =\frac{0.095}{3}$ = 0.032 mA

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Read 2 more answers

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The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

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Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

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Case for the particle in the Encke Division:

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$T = \sqrt{(133370 Km)^{3}}$

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$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

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For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

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$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

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3 years ago