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GRADE 8 PRACTICE

Physics Problems: Magnetic phenomena, Grade 8

Magnetic fields, electromagnets and motor.

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 #61

Magnetic field and electromagnet

What concept does the description correspond to: “A form of matter that acts on magnets, moving charges, and current-carrying conductors.»?

Answer options

  1. \text{Magnetic field}
  2. \text{Magnetic field of current}
  3. \text{Electromagnet}
  4. \text{Capillary action}
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Given
\text{Description}
\text{A form of matter that acts on magnets, moving charges, and current-carrying conductors.}
Find

\text{name concepts}

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

    Magnetic lines outside a permanent magnet are directed from the north pole to the south pole and do not intersect.

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

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

Position in this set: 2 from 18

Problem #62

Magnetic field and electromagnet

What concept does the description correspond to: “Electric current creates a magnetic field around the conductor, the direction of which is related to the direction of the current.»?

Answer options

  1. \text{Magnetic field of current}
  2. \text{Magnetic field}
  3. \text{Electromagnet}
  4. \text{Capillary action}
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Given
\text{Description}
\text{Electric current creates a magnetic field around the conductor, the direction of which is related to the direction of the current.}
Find

\text{name concepts}

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

    A coil's individual turn fields add, forming a field similar to a bar magnet's.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Magnetic field of current}

Answer\text{Concept} = \text{Magnetic field of current}

Position in this set: 3 from 18

Problem #63

Magnetic field

What change, while maintaining the size of the coil, usually increases its magnetic effect?

Answer options

  1. \text{increase current and Number of turns}
  2. \text{reduce current and remove core}
  3. \text{Open chain}
  4. \text{Replace wire insulator}
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Given
\text{Reel}
\text{dimensions permanent}
Find

\text{way gain}

  1. \text{Fields turns fold up}
  2. \text{larger current enhances field everyone turn}\text{more current and turns}

Answer\text{increase current and Number of turns}

Position in this set: 4 from 18

Problem #64

Magnetic field action and motor

What concept does the description correspond to: “A magnetic field exerts a force on a current-carrying conductor if the current is not directed along the field lines.»?

Answer options

  1. \text{Action of the field on the conductor}
  2. \text{Electric motor}
  3. \text{Earth's magnetic field}
  4. \text{Capillary action}
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Given
\text{Description}
\text{A magnetic field exerts a force on a current-carrying conductor if the current is not directed along the field lines.}
Find

\text{name concepts}

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

    Use the left-hand rule with current direction and magnetic flux density direction to find the force direction.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Action of the field on the conductor}

Answer\text{Concept} = \text{Action of the field on the conductor}

Position in this set: 5 from 18

Problem #65

Magnetic field action and motor

What concept does the description correspond to: “A device that converts electrical energy into mechanical energy due to the action of a magnetic field on a current.»?

Answer options

  1. \text{Electric motor}
  2. \text{Action of the field on the conductor}
  3. \text{Earth's magnetic field}
  4. \text{Capillary action}
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Given
\text{Description}
\text{A device that converts electrical energy into mechanical energy due to the action of a magnetic field on a current.}
Find

\text{name concepts}

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

    Forces on opposite sides of the loop produce torque; switching current direction maintains rotation.

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

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

Position in this set: 6 from 18

Problem #66

Electric motor

A conductor 0.1 m long is perpendicular to a field of induction 0.2 T. A current of 3 A flows through it. Find the force.

Answer options

  1. 0{,}06\,\text{Н}
  2. 0{,}006\,\text{Н}
  3. 0{,}6\,\text{Н}
  4. 6\,\text{Н}
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Given
B
0{,}2\,\text{Тл}
I
3\,\text{А}
l
0{,}1\,\text{м}
Find

F

  1. F=BIlF=0{,}2\cdot3\cdot0{,}1\,\text{Н}F = 0{,}06\,\text{Н}

Answer0{,}06\,\text{Н}

Position in this set: 7 from 18

Problem #193

Magnetic field and electromagnet

On core reeled 4 identical layers by 50 turns. How much total turns has electromagnet?

Answer options

  1. 200\,\text{витков}
  2. 400\,\text{витков}
  3. 100\,\text{витков}
  4. 600\,\text{витков}
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Given
n
4\,\text{слоёв}
k
50\,\text{витков}
Find

N

  1. N=nkN=4 \cdot 50N=200\,\text{витков}

    Add the same number of turns in all layers.

Answer200\,\text{витков}

Position in this set: 8 from 18

Problem #194

Magnetic field and electromagnet

On core reeled 6 identical layers by 40 turns. How much total turns has electromagnet?

Answer options

  1. 480\,\text{витков}
  2. 240\,\text{витков}
  3. 120\,\text{витков}
  4. 720\,\text{витков}
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Given
n
6\,\text{слоёв}
k
40\,\text{витков}
Find

N

  1. N=nkN=6 \cdot 40N=240\,\text{витков}

    Add the same number of turns in all layers.

Answer240\,\text{витков}

Position in this set: 9 from 18

Problem #195

Magnetic field and electromagnet

For identical cores compare work numbers turns on current NI. First reel has 200 turns at current 0{,}6 A. Which current needed reel from 400 turns, to NI Matched?

Answer options

  1. 0{,}6\,\text{А}
  2. 0{,}3\,\text{А}
  3. 0{,}15\,\text{А}
  4. 0{,}9\,\text{А}
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Given
N_1
200\,\text{витков}
I_1
0{,}6\,\text{А}
N_2
400\,\text{витков}
Find

I_2

  1. a=N_1I_1a=200 \cdot 0{,}6a=120\,\text{А·витков}

    We use the specified comparison criterion.

  2. I_2=\frac{a}{N_2}I_2=\frac{120}{400}I_2=0{,}3\,\text{А}

    Twice the number of turns requires half the current for the same NI.

Answer0{,}3\,\text{А}

Position in this set: 10 from 18

Problem #196

Magnetic field and electromagnet

For identical cores compare work numbers turns on current NI. First reel has 300 turns at current 0{,}8 A. Which current needed reel from 600 turns, to NI Matched?

Answer options

  1. 0{,}8\,\text{А}
  2. 0{,}2\,\text{А}
  3. 0{,}4\,\text{А}
  4. 1{,}2\,\text{А}
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Given
N_1
300\,\text{витков}
I_1
0{,}8\,\text{А}
N_2
600\,\text{витков}
Find

I_2

  1. a=N_1I_1a=300 \cdot 0{,}8a=240\,\text{А·витков}

    We use the specified comparison criterion.

  2. I_2=\frac{a}{N_2}I_2=\frac{240}{600}I_2=0{,}4\,\text{А}

    Twice the number of turns requires half the current for the same NI.

Answer0{,}4\,\text{А}

Position in this set: 11 from 18

Problem #197

Magnetic field and electromagnet

Electromagnet has 300 turns and resistance 6 Ohm, connected K 12 B. Needed save work NI, replacing reel on 600 turns. Find required current new coils.

Answer options

  1. 2\,\text{А}
  2. 0{,}5\,\text{А}
  3. 1\,\text{А}
  4. 3\,\text{А}
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Given
N
300\,\text{витков}
R
6\,\text{Ом}
U
12\,\text{В}
M
600\,\text{витков}
Find

J

  1. I=\frac{U}{R}I=\frac{12}{6}I=2\,\text{А}

    The current of the original coil is determined by its resistance.

  2. a=NIa=300 \cdot 2a=600\,\text{А·витков}

    We find the original product of turns and current.

  3. J=\frac{a}{M}J=\frac{600}{600}J=1\,\text{А}

    For the new coil, we save this product as required by the condition.

Answer1\,\text{А}

Position in this set: 12 from 18

Problem #198

Magnetic field and electromagnet

Electromagnet has 400 turns and resistance 8 Ohm, connected K 24 B. Needed save work NI, replacing reel on 800 turns. Find required current new coils.

Answer options

  1. 3\,\text{А}
  2. 0{,}75\,\text{А}
  3. 4{,}5\,\text{А}
  4. 1{,}5\,\text{А}
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Given
N
400\,\text{витков}
R
8\,\text{Ом}
U
24\,\text{В}
M
800\,\text{витков}
Find

J

  1. I=\frac{U}{R}I=\frac{24}{8}I=3\,\text{А}

    The current of the original coil is determined by its resistance.

  2. a=NIa=400 \cdot 3a=1200\,\text{А·витков}

    We find the original product of turns and current.

  3. J=\frac{a}{M}J=\frac{1200}{800}J=1{,}5\,\text{А}

    For the new coil, we save this product as required by the condition.

Answer1{,}5\,\text{А}

Position in this set: 13 from 18

Problem #199

Magnetic field action and motor

Engine develops useful power 40 W B current 30 s. Find useful work.

Answer options

  1. 2400\,\text{Дж}
  2. 1200\,\text{Дж}
  3. 600\,\text{Дж}
  4. 3600\,\text{Дж}
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Given
P
40\,\text{Вт}
t
30\,\text{с}
Find

A

  1. A=PtA=40 \cdot 30A=1200\,\text{Дж}

    The rotation of the motor allows mechanical work to be done.

Answer1200\,\text{Дж}

Position in this set: 14 from 18

Problem #200

Magnetic field action and motor

Engine develops useful power 60 W B current 20 s. Find useful work.

Answer options

  1. 2400\,\text{Дж}
  2. 600\,\text{Дж}
  3. 1200\,\text{Дж}
  4. 3600\,\text{Дж}
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Given
P
60\,\text{Вт}
t
20\,\text{с}
Find

A

  1. A=PtA=60 \cdot 20A=1200\,\text{Дж}

    The rotation of the motor allows mechanical work to be done.

Answer1200\,\text{Дж}

Position in this set: 15 from 18

Problem #201

Magnetic field action and motor

Engine Under tension 12 B consumes current 5 A. His Useful Mechanical power 48 W. Find EFFICIENCY B percent.

Answer options

  1. 160\,\text{\%}
  2. 40\,\text{\%}
  3. 80\,\text{\%}
  4. 240\,\text{\%}
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Given
U
12\,\text{В}
I
5\,\text{А}
P
48\,\text{Вт}
Find

\eta

  1. P_{\text{вх}}=UIP_{\text{вх}}=12 \cdot 5P_{\text{вх}}=60\,\text{Вт}

    Find the consumed electrical power.

  2. \eta=\frac{P}{P_{\text{вх}}}\cdot100\eta=\frac{48}{60} \cdot 100\eta=80\,\text{\%}

    Compare mechanical and electrical power.

Answer80\,\text{\%}

Position in this set: 16 from 18

Problem #202

Magnetic field action and motor

Engine Under tension 24 B consumes current 4 A. His Useful Mechanical power 72 W. Find EFFICIENCY B percent.

Answer options

  1. 150\,\text{\%}
  2. 37{,}5\,\text{\%}
  3. 225\,\text{\%}
  4. 75\,\text{\%}
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Given
U
24\,\text{В}
I
4\,\text{А}
P
72\,\text{Вт}
Find

\eta

  1. P_{\text{вх}}=UIP_{\text{вх}}=24 \cdot 4P_{\text{вх}}=96\,\text{Вт}

    Find the consumed electrical power.

  2. \eta=\frac{P}{P_{\text{вх}}}\cdot100\eta=\frac{72}{96} \cdot 100\eta=75\,\text{\%}

    Compare mechanical and electrical power.

Answer75\,\text{\%}

Position in this set: 17 from 18

Problem #203

Magnetic field action and motor

Engine eats from 24 B at current 5 A. EFFICIENCY 75 %. For 20 with he evenly raises Load weight 30 kg. Find height rise, g = 10 N/kg.

Answer options

  1. 12\,\text{м}
  2. 3\,\text{м}
  3. 18\,\text{м}
  4. 6\,\text{м}
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Given
U
24\,\text{В}
I
5\,\text{А}
e
75\,\text{\%}
t
20\,\text{с}
m
30\,\text{кг}
g
10\,\text{Н/кг}
Find

h

  1. E=UItE=24 \cdot 5 \cdot 20E=2400\,\text{Дж}

    Finding the energy consumed.

  2. A=E\frac{\eta}{100}A=\frac{2400 \cdot 75}{100}A=1800\,\text{Дж}

    We take into account the conversion efficiency.

  3. h=\frac{A}{mg}h=\frac{1800}{(30 \cdot 10)}h=6\,\text{м}

    Useful work is equal to mgh.

Answer6\,\text{м}

Position in this set: 18 from 18

Problem #204

Magnetic field action and motor

Engine eats from 30 B at current 8 A. EFFICIENCY 80 %. For 25 with he evenly raises Load weight 40 kg. Find height rise, g = 10 N/kg.

Answer options

  1. 12\,\text{м}
  2. 24\,\text{м}
  3. 6\,\text{м}
  4. 36\,\text{м}
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Given
U
30\,\text{В}
I
8\,\text{А}
e
80\,\text{\%}
t
25\,\text{с}
m
40\,\text{кг}
g
10\,\text{Н/кг}
Find

h

  1. E=UItE=30 \cdot 8 \cdot 25E=6000\,\text{Дж}

    Finding the energy consumed.

  2. A=E\frac{\eta}{100}A=\frac{6000 \cdot 80}{100}A=4800\,\text{Дж}

    We take into account the conversion efficiency.

  3. h=\frac{A}{mg}h=\frac{4800}{(40 \cdot 10)}h=12\,\text{м}

    Useful work is equal to mgh.

Answer12\,\text{м}

The theory behind these problems

Work through the topic if a problem will not give in.

Magnetic field and electromagnetPermanent magnets, the magnetic field of current, coils and uses of electromagnets.Magnetic field action and motorThe Earth's field, the left-hand rule and the principle of the electric motor.

Formulas by topic

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ForceF · НElectric currentI · А