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Mariana [72]
2 years ago
15

How many turns should a solenoid of cross-sectional area 3.3×10−2 m2 and length 0.30 m have if its inductance is to be 47 mh ?

Physics
1 answer:
Nikitich [7]2 years ago
4 0
The inductance of a solenoid is given by
L=\mu  \frac{N^2}{l} A
where
\mu = 12.56 \cdot 10^{-7}NA^{-2} is the magnetic permittivity
N is the number of turns of the solenoid
l is its lenght
A is its cross-sectional area

By re-arranging the formula and replacing the numbers, we get the number of turns of the solenoid:
N= \sqrt{ \frac{lL}{\mu A} }= \sqrt{ \frac{(4.7\cdot 10^{-3}H)(0.3 m)}{(12.56\cdot 10^{-7}NA^{-2})(3.3\cdot10^{-2}m^2)} }  =583
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Answer:

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Explanation:

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The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.

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In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.

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