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The field winding of a dc electro-magnet is wound with 960 turns and has a resistance of 50 Ω. The excitation voltage is 230 V and the magnetic flux linking the coil is 5 mWb. The energy stored in the magnetic field is
- a)22 Joules
- b)11 Joules
- c)19 Joules
- d)29 Joules
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The field winding of a dc electro-magnet is wound with 960 turns and h...
Current through the field winding,
So, energy stored in magnetic field
So, energy stored in magnetic field
Most Upvoted Answer
The field winding of a dc electro-magnet is wound with 960 turns and h...
Ohms. If a current of 2.5 A is passed through the winding, what is the magnetic field strength produced by the magnet?
We can use the formula for the magnetic field strength (H) produced by a dc electromagnet:
H = (N * I) / l
where N is the number of turns in the winding, I is the current passing through the winding, and l is the length of the magnetic path.
In this case, we are given that N = 960 turns, I = 2.5 A, and we need to find l. We can use Ohm's Law to find the voltage drop across the winding:
V = I * R
where R is the resistance of the winding. Plugging in the given values, we get:
V = 2.5 A * 50 ohms = 125 V
Now, we can use the formula for the voltage drop across a resistor:
V = I * R
where V is the voltage drop across the winding, I is the current passing through the winding, and R is the resistance of the winding. Rearranging this formula, we get:
R = V / I
Plugging in the values we just found, we get:
l = V / (N * I) = 125 V / (960 turns * 2.5 A) = 0.0521 m
Now we can plug all of our values into the formula for H:
H = (N * I) / l = (960 turns * 2.5 A) / 0.0521 m = 46,312 A/m
Therefore, the magnetic field strength produced by the magnet is 46,312 A/m.
We can use the formula for the magnetic field strength (H) produced by a dc electromagnet:
H = (N * I) / l
where N is the number of turns in the winding, I is the current passing through the winding, and l is the length of the magnetic path.
In this case, we are given that N = 960 turns, I = 2.5 A, and we need to find l. We can use Ohm's Law to find the voltage drop across the winding:
V = I * R
where R is the resistance of the winding. Plugging in the given values, we get:
V = 2.5 A * 50 ohms = 125 V
Now, we can use the formula for the voltage drop across a resistor:
V = I * R
where V is the voltage drop across the winding, I is the current passing through the winding, and R is the resistance of the winding. Rearranging this formula, we get:
R = V / I
Plugging in the values we just found, we get:
l = V / (N * I) = 125 V / (960 turns * 2.5 A) = 0.0521 m
Now we can plug all of our values into the formula for H:
H = (N * I) / l = (960 turns * 2.5 A) / 0.0521 m = 46,312 A/m
Therefore, the magnetic field strength produced by the magnet is 46,312 A/m.
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The field winding of a dc electro-magnet is wound with 960 turns and has a resistance of 50 Ω. The excitation voltage is 230 V and the magnetic flux linking the coil is 5 mWb. The energy stored in the magnetic field isa)22 Joulesb)11 Joulesc)19 Joulesd)29 JoulesCorrect answer is option 'B'. Can you explain this answer?
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The field winding of a dc electro-magnet is wound with 960 turns and has a resistance of 50 Ω. The excitation voltage is 230 V and the magnetic flux linking the coil is 5 mWb. The energy stored in the magnetic field isa)22 Joulesb)11 Joulesc)19 Joulesd)29 JoulesCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The field winding of a dc electro-magnet is wound with 960 turns and has a resistance of 50 Ω. The excitation voltage is 230 V and the magnetic flux linking the coil is 5 mWb. The energy stored in the magnetic field isa)22 Joulesb)11 Joulesc)19 Joulesd)29 JoulesCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The field winding of a dc electro-magnet is wound with 960 turns and has a resistance of 50 Ω. The excitation voltage is 230 V and the magnetic flux linking the coil is 5 mWb. The energy stored in the magnetic field isa)22 Joulesb)11 Joulesc)19 Joulesd)29 JoulesCorrect answer is option 'B'. Can you explain this answer?.
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