Suppose that the current in the solenoid is i(t. within the solenoid, but far from its ends, what is the magnetic field b(t due
to this current?
2 answers:
The magnetic field b(t) due to this current is
Suppose that the current in the solenoid is i(t). Within the solenoid, but far from its ends, what is the magnetic field B(t) due to this current?
The magnetic field is the area around a magnet where there is magnetic force. Moving electric charges can make magnetic fields. Then a solenoid is the long coil of wire wrapped in many turns. When a current passes through, it creates a nearly uniform magnetic field inside.
Solenoids can convert electric current to mechanical action, and so are very commonly used as switches
Even small solenoids can exert forces of a few newtons.
If we look through the solenoid far from ends, we can use Ampere's law to calculate the field strength. The magnetic field (deep) within the solenoid has a uniform value B, and outside the coils has value zero.
The magnetic field within a solenoid depends upon the current and density of turns. The magnetic field B(t) due to this current is . This is derived from Ampere's law used to calculate the strength of magnetic field.
Where is number of coils per meter and
is current through wire.
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Answer:
a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.
b) Average power
P(w)= 1062.07 [w]
P(hp)=1.42 [hp]
c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.
Explanation:
First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)
Work:
Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )
Converting from Joules to Kcals:
Now lets take into account the efficiency of the human body (20%)
2.537 ---> 20%
x ---> 100%
So the student is consuming 12.685 KCals each time he runs up the stairs.
Now,
1 g --> 9 Kcals
1000 g --> 9000KCals
Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:
He must run up the stairs 709.5 times, to burn 1 Kg of fat.
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For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)
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