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GRADE 8 PRACTICE

Physics Problems: Electromagnetic induction, Grade 8

Induction current, Lenz's rule and energy production.

18 problems across 4 sub-topics
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Problem statements

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18 problems
Position in this set: 1 from 18

Problem #67

Electromagnetic induction and Lenz's law

What concept does the description correspond to: “The production of an emf when magnetic flux through a loop changes. Induced current flows if the conducting loop is closed.»?

Answer options

  1. \text{Electromagnetic induction}
  2. \text{Magnetic flux}
  3. \text{Lenz's law}
  4. \text{Capillary action}
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Given
\text{Description}
\text{The production of an emf when magnetic flux through a loop changes. Induced current flows if the conducting loop is closed.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Current can arise when a magnet moves relative to a coil, the current in a neighbouring coil changes, or a loop rotates.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Electromagnetic induction}

Answer\text{Concept} = \text{Electromagnetic induction}

Position in this set: 2 from 18

Problem #68

Electromagnetic induction and Lenz's law

What concept does the description correspond to: “A value characterizing the magnetic field passing through the selected surface.»?

Answer options

  1. \text{Magnetic flux}
  2. \text{Electromagnetic induction}
  3. \text{Lenz's law}
  4. \text{Capillary action}
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Given
\text{Description}
\text{A value characterizing the magnetic field passing through the selected surface.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Flux changes when the field, area or orientation of the loop relative to the magnetic field lines changes.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Magnetic flux}

Answer\text{Concept} = \text{Magnetic flux}

Position in this set: 3 from 18

Problem #69

Electromagnetic induction

The north pole of the magnet is brought closer to a closed conducting coil. What is the direction of the induced current field at the near end of the coil?

Answer options

  1. \text{Middle end becomes northern and pushes away magnet}
  2. \text{Middle end becomes southern and attracts magnet}
  3. \text{Field always disappears}
  4. \text{direction not connected with movement magnet}
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Given
\text{Change}
\text{northern pole approaching}
Find

\text{answer fields coils}

  1. \text{Magnetic flux through reel increases}
  2. \text{By rule Lenz reel opposes growth flow}\text{pushes away northern pole}

Answer\text{neighbor end becomes northern}

Position in this set: 4 from 18

Problem #70

Generators and electrical energy transmission

What concept does the description correspond to: “Energy is transferred through wires at high voltage to reduce current and heat loss.»?

Answer options

  1. \text{Electrical energy transmission}
  2. \text{Electric generator}
  3. \text{Transformer}
  4. \text{Capillary action}
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Given
\text{Description}
\text{Energy is transferred through wires at high voltage to reduce current and heat loss.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The transformer reduces the voltage to the operating value specified for the equipment. Household mains voltage remains dangerous for humans.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Electrical energy transmission}

Answer\text{Concept} = \text{Electrical energy transmission}

Position in this set: 5 from 18

Problem #71

Generators and electrical energy transmission

What concept does the description correspond to: “A device that converts mechanical rotational energy into electrical energy through electromagnetic induction.»?

Answer options

  1. \text{Electric generator}
  2. \text{Electrical energy transmission}
  3. \text{Transformer}
  4. \text{Capillary action}
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Given
\text{Description}
\text{A device that converts mechanical rotational energy into electrical energy through electromagnetic induction.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Rotation of a coil or magnet periodically changes the flux and creates an alternating voltage.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Electric generator}

Answer\text{Concept} = \text{Electric generator}

Position in this set: 6 from 18

Problem #72

Energy transfer

The primary winding has 100 turns and is connected to an alternating voltage with an effective value of 22 V. The secondary has 1000 turns. Find the effective value of the secondary voltage of an ideal transformer.

Answer options

  1. 220\,\text{В}
  2. 2{,}2\,\text{В}
  3. 22\,\text{В}
  4. 2200\,\text{В}
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Given
N_1
100
N_2
1000
U_1
22\,\text{В}
Find

U_2

  1. \frac{U_1}{U_2}=\frac{N_1}{N_2}
  2. U_2=U_1\frac{N_2}{N_1}U_2=22\cdot\frac{1000}{100}\,\text{В}U_2 = 220\,\text{В}

Answer220\,\text{В}

Position in this set: 7 from 18

Problem #205

Electromagnetic induction and Lenz's law

When one turnover magnet sensor issues 2 impulses. magnet committed 30 rpm. How much impulses registered?

Answer options

  1. 60\,\text{импульсов}
  2. 120\,\text{импульсов}
  3. 30\,\text{импульсов}
  4. 180\,\text{импульсов}
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Given
n
2\,\text{импульсов/оборот}
N
30\,\text{оборотов}
Find

K

  1. K=nNK=2 \cdot 30K=60\,\text{импульсов}

    Changes in the magnetic field are repeated with each revolution.

Answer60\,\text{импульсов}

Position in this set: 8 from 18

Problem #206

Electromagnetic induction and Lenz's law

When one turnover magnet sensor issues 4 impulses. magnet committed 20 rpm. How much impulses registered?

Answer options

  1. 160\,\text{импульсов}
  2. 80\,\text{импульсов}
  3. 40\,\text{импульсов}
  4. 240\,\text{импульсов}
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Given
n
4\,\text{импульсов/оборот}
N
20\,\text{оборотов}
Find

K

  1. K=nNK=4 \cdot 20K=80\,\text{импульсов}

    Changes in the magnetic field are repeated with each revolution.

Answer80\,\text{импульсов}

Position in this set: 9 from 18

Problem #207

Electromagnetic induction and Lenz's law

Sensor registers 120 impulses for 10 with, by 2 impulses on turnover magnet. Find number rpm B second.

Answer options

  1. 12\,\text{об/с}
  2. 6\,\text{об/с}
  3. 3\,\text{об/с}
  4. 18\,\text{об/с}
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Given
K
120\,\text{импульсов}
t
10\,\text{с}
n
2\,\text{импульсов/оборот}
Find

f

  1. N=\frac{K}{n}N=\frac{120}{2}N=60\,\text{оборотов}

    By the number of pulses we determine the number of revolutions.

  2. f=\frac{N}{t}f=\frac{60}{10}f=6\,\text{об/с}

    We divide the number of revolutions by the duration of observation.

Answer6\,\text{об/с}

Position in this set: 10 from 18

Problem #208

Electromagnetic induction and Lenz's law

Sensor registers 240 impulses for 20 with, by 4 impulses on turnover magnet. Find number rpm B second.

Answer options

  1. 6\,\text{об/с}
  2. 1{,}5\,\text{об/с}
  3. 3\,\text{об/с}
  4. 9\,\text{об/с}
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Given
K
240\,\text{импульсов}
t
20\,\text{с}
n
4\,\text{импульсов/оборот}
Find

f

  1. N=\frac{K}{n}N=\frac{240}{4}N=60\,\text{оборотов}

    By the number of pulses we determine the number of revolutions.

  2. f=\frac{N}{t}f=\frac{60}{20}f=3\,\text{об/с}

    We divide the number of revolutions by the duration of observation.

Answer3\,\text{об/с}

Position in this set: 11 from 18

Problem #209

Electromagnetic induction and Lenz's law

Generator for 20 with passed on load current with average power 30 W and voltage 12 B. Mechanical work rotation amounted to 1000 J. Find EFFICIENCY generator; for calculation energy use given average power.

Answer options

  1. 120\,\text{\%}
  2. 30\,\text{\%}
  3. 60\,\text{\%}
  4. 180\,\text{\%}
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Given
t
20\,\text{с}
P
30\,\text{Вт}
U
12\,\text{В}
A
1000\,\text{Дж}
Find

\eta

  1. I=\frac{P}{U}I=\frac{30}{12}I=2{,}5\,\text{А}

    For the specified mode, we find the equivalent current in terms of power.

  2. E=PtE=30 \cdot 20E=600\,\text{Дж}

    Electrical energy is obtained through mechanical work.

  3. \eta=\frac{E}{A}\cdot100\eta=\frac{600}{1000} \cdot 100\eta=60\,\text{\%}

    Lenz's rule is consistent with the need to expend work on generation.

Answer60\,\text{\%}

Position in this set: 12 from 18

Problem #210

Electromagnetic induction and Lenz's law

Generator for 30 with passed on load current with average power 40 W and voltage 24 B. Mechanical work rotation amounted to 2000 J. Find EFFICIENCY generator; for calculation energy use given average power.

Answer options

  1. 120\,\text{\%}
  2. 30\,\text{\%}
  3. 180\,\text{\%}
  4. 60\,\text{\%}
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Given
t
30\,\text{с}
P
40\,\text{Вт}
U
24\,\text{В}
A
2000\,\text{Дж}
Find

\eta

  1. I=\frac{P}{U}I=\frac{40}{24}I=1{,}66667\,\text{А}

    For the specified mode, we find the equivalent current in terms of power.

  2. E=PtE=40 \cdot 30E=1200\,\text{Дж}

    Electrical energy is obtained through mechanical work.

  3. \eta=\frac{E}{A}\cdot100\eta=\frac{1200}{2000} \cdot 100\eta=60\,\text{\%}

    Lenz's rule is consistent with the need to expend work on generation.

Answer60\,\text{\%}

Position in this set: 13 from 18

Problem #211

Generators and electrical energy transmission

By lines transmitted power 12000 W Under tension 2000 B. Find current; use P = UI.

Answer options

  1. 12\,\text{А}
  2. 6\,\text{А}
  3. 3\,\text{А}
  4. 18\,\text{А}
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Given
P
12000\,\text{Вт}
U
2000\,\text{В}
Find

I

  1. I=\frac{P}{U}I=\frac{12000}{2000}I=6\,\text{А}

    For the same power, increasing the voltage reduces the current.

Answer6\,\text{А}

Position in this set: 14 from 18

Problem #212

Generators and electrical energy transmission

By lines transmitted power 24000 W Under tension 3000 B. Find current; use P = UI.

Answer options

  1. 16\,\text{А}
  2. 4\,\text{А}
  3. 8\,\text{А}
  4. 24\,\text{А}
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Given
P
24000\,\text{Вт}
U
3000\,\text{В}
Find

I

  1. I=\frac{P}{U}I=\frac{24000}{3000}I=8\,\text{А}

    For the same power, increasing the voltage reduces the current.

Answer8\,\text{А}

Position in this set: 15 from 18

Problem #213

Generators and electrical energy transmission

transmitted power 10000 W, voltage 1000 V, resistance wires 2 Ohm. Find power losses I²R.

Answer options

  1. 400\,\text{Вт}
  2. 100\,\text{Вт}
  3. 200\,\text{Вт}
  4. 600\,\text{Вт}
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Given
P
10000\,\text{Вт}
U
1000\,\text{В}
R
2\,\text{Ом}
Find

P_{\text{пот}}

  1. I=\frac{P}{U}I=\frac{10000}{1000}I=10\,\text{А}

    Finding the transmission current.

  2. P_{\text{пот}}=I^2RP_{\text{пот}}={10}^{2} \cdot 2P_{\text{пот}}=200\,\text{Вт}

    Heating the wires consumes part of the transmitted energy.

Answer200\,\text{Вт}

Position in this set: 16 from 18

Problem #214

Generators and electrical energy transmission

transmitted power 24000 W, voltage 2000 V, resistance wires 1 Ohm. Find power losses I²R.

Answer options

  1. 288\,\text{Вт}
  2. 72\,\text{Вт}
  3. 432\,\text{Вт}
  4. 144\,\text{Вт}
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Given
P
24000\,\text{Вт}
U
2000\,\text{В}
R
1\,\text{Ом}
Find

P_{\text{пот}}

  1. I=\frac{P}{U}I=\frac{24000}{2000}I=12\,\text{А}

    Finding the transmission current.

  2. P_{\text{пот}}=I^2RP_{\text{пот}}={12}^{2} \cdot 1P_{\text{пот}}=144\,\text{Вт}

    Heating the wires consumes part of the transmitted energy.

Answer144\,\text{Вт}

Position in this set: 17 from 18

Problem #215

Generators and electrical energy transmission

By one lines resistance 2 Ohm transmit power 10000 W. Voltage promoted with 1000 up to 2000 B. On how much watt decreased losses? Use P = UI and power losses I²R.

Answer options

  1. 300\,\text{Вт}
  2. 75\,\text{Вт}
  3. 450\,\text{Вт}
  4. 150\,\text{Вт}
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Given
R
2\,\text{Ом}
P
10000\,\text{Вт}
U_1
1000\,\text{В}
U_2
2000\,\text{В}
Find

\Delta P

  1. I_1=\frac{P}{U_1}I_1=\frac{10000}{1000}I_1=10\,\text{А}

    Current before voltage rises.

  2. I_2=\frac{P}{U_2}I_2=\frac{10000}{2000}I_2=5\,\text{А}

    Current after voltage increase.

  3. \Delta P=(I_1^2-I_2^2)R\Delta P=({10}^{2} - {5}^{2}) \cdot 2\Delta P=150\,\text{Вт}

    When the voltage is doubled, the losses are reduced fourfold.

Answer150\,\text{Вт}

Position in this set: 18 from 18

Problem #216

Generators and electrical energy transmission

By one lines resistance 3 Ohm transmit power 12000 W. Voltage promoted with 1200 up to 2400 B. On how much watt decreased losses? Use P = UI and power losses I²R.

Answer options

  1. 225\,\text{Вт}
  2. 450\,\text{Вт}
  3. 112{,}5\,\text{Вт}
  4. 675\,\text{Вт}
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Given
R
3\,\text{Ом}
P
12000\,\text{Вт}
U_1
1200\,\text{В}
U_2
2400\,\text{В}
Find

\Delta P

  1. I_1=\frac{P}{U_1}I_1=\frac{12000}{1200}I_1=10\,\text{А}

    Current before voltage rises.

  2. I_2=\frac{P}{U_2}I_2=\frac{12000}{2400}I_2=5\,\text{А}

    Current after voltage increase.

  3. \Delta P=(I_1^2-I_2^2)R\Delta P=({10}^{2} - {5}^{2}) \cdot 3\Delta P=225\,\text{Вт}

    When the voltage is doubled, the losses are reduced fourfold.

Answer225\,\text{Вт}

The theory behind these problems

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Electromagnetic induction and Lenz's lawMagnetic flux and the origin and direction of induced current.Generators and electrical energy transmissionConversion of mechanical energy and transmission with changes in voltage.

Formulas by topic

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Electrical voltageU · В