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4 years ago

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You're an electrical engineer designing an alternator (the generator that charges a car's battery). Mechanical engineers specify

a 10-cm-diameter rotating coil, and you determine that you can fit 250 turns in this coil. To charge a 12-Vbattery, you need a peak output of 15V when the alternator is rotating at 1000 rpm. What do you specify for the alternator's magnetic field?

1 answer:

____ [38]4 years ago

6 0

Answer:

B=0.07299 T( required magnitude of Magnetic field)

Explanation:

Given:

diameter of coil= 10 cm implies radius=5 cm

emf induced= 14V

rmp N= 1000

therefore, $\omega=\frac{2\times \pi\times N}{60}$

put values $\omega=\frac{2\times \pi\times 1000}{60}$

=104.71 rad/sec

The area of coil is given by

A= $\pi\times r^{2}$

A= $\pi\times0.05^{2}$

= $7.85\times10^{3}m^{2}$

The expression of peak Emf induced in the rotating coil is

is given by

$\epsilon_{max}=NA\omega B$

rearranging the expression to find B

$B= \frac{\epsilon_{max}}{NA\omega }$

$B= \frac{15}{250\times7.85\times10^{-3}\times104.71 }$

on calculating we get

B=0.07299 T( required magnitude of Magnetic field)

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3 years ago

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

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Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

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According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

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