Respuesta :
Answer:
3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled
4. Doubling the voltage, doubles the strength of the electromagnet
5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips
The number of paper clips a 7.5 V battery would pick is 59 paperclips
6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips
For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips
Explanation:
3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I
∴ MMF = N × I
When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled
4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have
V = I × R
Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet
5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3.[tex]\bar 6[/tex]
The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips ≈ 28 paper clips
The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7.[tex]\bar 3[/tex]
The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips
6. The slope calculated from a start point of approximately 0.4 V, is given as follows;
The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13
Therefore, for the 25-coil electromagnet, the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips
The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5
Therefore, for the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips
3. When the number of revolutions, N, is twice, the electromagnet's strength is doubled as well.
4. Doubling the voltage doubles the electromagnet's strength.
5. The number of paper clips selected by a 7.5 V battery is 59.
6. For a 50-coil electromagnet, a 1 V battery would pull up around 7 paperclips on average. The average number of paper clips picked by a 1 V battery for a 50-coil electromagnet is 16 paperclips.
What is electromagnet?
The magnetic field of an electromagnet is created by an electric current. Electromagnets are caused from conducting wire wriggled into a coil.
A magnetic field is created in the hole by a current passing through the wire.
The magnetic field production is increased by;
1) increasing the number of turns
2) increase the area of the loop
3) by moving the magnet faster
Answer for the following given options as follows;
3. When the number of revolutions, N, is twice, the electromagnet's strength is doubled as well. Because the no of coils in the circuit is directly proportional to the strength of electromagnet.
4. From the ohm's law principal doubling the voltage twice the current and hence the electromagnet's strength for a given constant resistance in the coil increases.
5. For the 25-coil electromagnet, the average slope is ;
[tex]\rm m_{avg}= \frac{23-12}{6-3} \\\\ m_{avg}= 3[/tex]
The number of paper clips selected by a 7.5 V battery is;
[tex]\rm n = 12+ (7.5-+3)\times \frac{11}{3}\\\\n = 28.5[/tex]
For the 50-coil electromagnet, the average slope;
[tex]\rm (m_{avg})_{50}= \frac{48-26}{6-3} \\\\\ \rm (m_{avg})_{50}= 7[/tex]
The number of paper clips selected by a 7.5 V battery is;
[tex]\rm N = 26+(7.5-3) \times \frac{22}{3} \\\\ N= 59[/tex]
The following is the slope determined from a starting position of around 0.4 V:
For the 25-coil electromagnet, the slope is;
[tex]\rm m_{25}=\frac{12-6}{3-0.4} \\\\\ m_{25}=\frac{6}{2.6} \\\\ m_{25}=\frac{30}{13}[/tex]
As a result, the average number of paper clips a 1 V battery would choose for the 25-coil electromagnet is;
[tex]\rm N_{25}=6+ (1-0.4)\times \frac{30}{13} \\\\ N_{25}=\frac{96}{13} \\\\ N_{25}= 7 \ paperclip[/tex]
For the 50-coil electromagnet, the slope is;
(26 - 13)/(3 - 0.4) = 5.
As a result, the average number of paper clips picked by a 1 V battery is;
[tex]\rm N'_{50}= 13+ (1-0.14)\times 5 \\\\ N'_{50}=16[/tex]
Hence,the average number of paper clips picked by a 1 V battery for a 50-coil electromagnet is 16 paperclips.
To learn more about the electromagnets, refers to the link:
https://brainly.com/question/23727978
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