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GRADE 11 PRACTICE

Physics Problems: Magnetism, Grade 11

Magnetic field forces and induction.

23 problems across 4 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

23 problems
Position in this set: 1 from 23

Problem #1

Magnetism: magnetic field forces

To a current-carrying conductor 5\,\text{А} and length 40\,\text{см} maximum force acting in a magnetic field 5\,\text{мН}. What is the magnetic field induction?

Answer options

  1. 0{,}005\,\text{Тл}
  2. 0{,}025\,\text{Тл}
  3. 0{,}0025\,\text{Тл}
  4. 0{,}00125\,\text{Тл}
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Given
I
5\,\text{А}
\ell
40\,\text{см}0{,}4\,\text{м}
F_{\text{м}}
5\,\text{мН}5\cdot 10^{-3}\,\text{Н}
Find

B

  1. F_{\text{м}} = I\ell B\sin\alpha

    The force is maximum at an angle of 90°, so the sine is equal to one.

  2. B = \dfrac{F_{\text{м}}}{I\ell}B = \dfrac{5\cdot 10^{-3}\,\text{Н}}{5\,\text{А}\cdot 0{,}4\,\text{м}}B = 2{,}5\cdot 10^{-3}\,\text{Тл} = 2{,}5\,\text{мТл}

AnswerB = 2{,}5\cdot 10^{-3}\,\text{Тл}

Position in this set: 2 from 23

Problem #2

Magnetism: magnetic field forces

What should be the angle between the conductor length 20\,\text{см} and the magnetic induction vector B=4\,\text{мТл}, so that with current 15\,\text{А} the strength was: a) 6\,\text{мН}, b) 18\,\text{мН}? Consider angles from 0° to 90°.

Answer options

  1. \left(60^{\circ};\ 30^{\circ}\right)
  2. \left(45^{\circ};\ 90^{\circ}\right)
  3. \left(30^{\circ};\ 60^{\circ}\right)
  4. \left(30^{\circ};\ \text{второй случай невозможен}\right)
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Given
\ell
20\,\text{см}0{,}2\,\text{м}
B
4\,\text{мТл}4\cdot 10^{-3}\,\text{Тл}
I
15\,\text{А}
F_1
6\cdot 10^{-3}\,\text{Н}
F_2
18\cdot 10^{-3}\,\text{Н}
Find

\alpha_1\alpha_2

  1. F = I\ell B\sin\alpha\ \Rightarrow\ \sin\alpha = \dfrac{F}{I\ell B}
  2. \sin\alpha_1 = \dfrac{6\cdot 10^{-3}}{15\cdot 0{,}2\cdot 4\cdot 10^{-3}}\alpha_1 = 0{,}5\ \Rightarrow\ \alpha_1 = 30^{\circ}
  3. \sin\alpha_2 = \dfrac{18\cdot 10^{-3}}{15\cdot 0{,}2\cdot 4\cdot 10^{-3}}1{,}5 > 1

    Sine cannot be greater than one: the maximum force under these conditions is 12 mN, so 18 mN is unattainable.

Answer\alpha_1 = 30^{\circ}\alpha_2\ \text{не существует}

Position in this set: 3 from 23

Problem #3

Magnetism: magnetic field forces

The electron moves at a speed 2\cdot10^{7}\,\text{м/с} perpendicular to the field with induction 40\,\text{мТл}. Determine the Lorentz force, the radius of the trajectory and the period of revolution.

Answer options

  1. \left(1{,}28 \cdot 10^{-13}\,\text{Н};\ 0{,}0028\,\text{м};\ 8{,}9 \cdot 10^{-10}\,\text{с}\right)
  2. \left(6{,}4 \cdot 10^{-14}\,\text{Н};\ 0{,}0014\,\text{м};\ 4{,}45 \cdot 10^{-10}\,\text{с}\right)
  3. \left(2{,}56 \cdot 10^{-13}\,\text{Н};\ 0{,}0056\,\text{м};\ 1{,}78 \cdot 10^{-9}\,\text{с}\right)
  4. \left(1{,}28 \cdot 10^{-12}\,\text{Н};\ 0{,}028\,\text{м};\ 8{,}9 \cdot 10^{-9}\,\text{с}\right)
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Given
B
40\,\text{мТл}40\cdot 10^{-3}\,\text{Тл}
v
2\cdot 10^{7}\,\text{м/с}
q
1{,}6\cdot 10^{-19}\,\text{Кл}
m
9{,}1\cdot 10^{-31}\,\text{кг}
\alpha
90^{\circ}
Find

F_{\text{Л}}RT

  1. F_{\text{Л}} = qvB\sin\alphaF_{\text{Л}} = 1{,}6\cdot 10^{-19}\cdot 2\cdot 10^{7}\cdot 40\cdot 10^{-3}F_{\text{Л}} = 1{,}28\cdot 10^{-13}\,\text{Н}
  2. F_{\text{Л}} = \dfrac{mv^{2}}{R}\ \Rightarrow\ R = \dfrac{mv^{2}}{F_{\text{Л}}}R = \dfrac{9{,}1\cdot 10^{-31}\cdot\left(2\cdot 10^{7}\right)^{2}}{1{,}28\cdot 10^{-13}}R = 0{,}0028\,\text{м} = 2{,}8\,\text{мм}

    The Lorentz force plays the role of a centripetal force.

  3. T = \dfrac{2\pi R}{v}T = \dfrac{2\cdot 3{,}14\cdot 0{,}0028}{2\cdot 10^{7}}T = 8{,}9\cdot 10^{-10}\,\text{с}

AnswerF_{\text{Л}} = 1{,}28\cdot 10^{-13}\,\text{Н}R = 0{,}0028\,\text{м}T = 8{,}9\cdot 10^{-10}\,\text{с}

Position in this set: 4 from 23

Problem #4

Magnetism: magnetic field forces

Homogeneous electric and magnetic fields are mutually perpendicular: E=0{,}5\,\text{кВ/м}, B=0{,}001\,\text{Тл}. The electron moves perpendicular to both fields. Find its speed and direction for uniform linear motion.

Answer options

  1. 5\cdot10^{5}\,\text{м/с}\ \text{вдоль электрического поля}
  2. v=5\cdot10^5\,\text{м/с},\quad \vec v\parallel\vec E\times\vec B
  3. 5\cdot10^{2}\,\text{м/с}\ \text{вдоль электрического поля}
  4. 2\cdot10^{6}\,\text{м/с}\ \text{вдоль магнитного поля}
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Given
E
0{,}5\,\tfrac{\text{кВ}}{\text{м}}500\,\tfrac{\text{В}}{\text{м}}
B
0{,}001\,\text{Тл}
Find

v

  1. F_{\text{Л}} = F_{\text{э}}

    The movement is uniform when the Lorentz force balances the electric force; We neglect the gravity force of the electron.

  2. qvB = qE\ \Rightarrow\ v = \dfrac{E}{B}v = \dfrac{500}{0{,}001}v = 5\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}

    Force equilibrium condition: E + v × B = 0. When v is perpendicular to both fields, the direction of velocity coincides with E × B.

Answerv=5\cdot10^{5}\,\text{м/с},\quad\vec v\parallel\vec E\times\vec B

Position in this set: 5 from 23

Problem #5

Magnetism: magnetic field forces

A proton flew into a magnetic field at a speed 500\,\text{км/с} perpendicular to the induction vector. Determine the orbital radius and orbital period at B=30\,\text{мТл}.

Answer options

  1. \left(0{,}348\,\text{м};\ 4{,}4 \cdot 10^{-6}\,\text{с}\right)
  2. \left(1{,}74\,\text{м};\ 2{,}2 \cdot 10^{-5}\,\text{с}\right)
  3. \left(0{,}174\,\text{м};\ 2{,}2 \cdot 10^{-6}\,\text{с}\right)
  4. \left(0{,}087\,\text{м};\ 1{,}1 \cdot 10^{-6}\,\text{с}\right)
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Given
v
500\,\tfrac{\text{км}}{\text{с}}500\cdot 10^{3}\,\text{м/с}
\alpha
90^{\circ}
B
30\cdot 10^{-3}\,\text{Тл}
q
1{,}6\cdot 10^{-19}\,\text{Кл}
m
1{,}67\cdot 10^{-27}\,\text{кг}
Find

RT

  1. qvB = \dfrac{mv^{2}}{R}\ \Rightarrow\ R = \dfrac{mv}{qB}R = \dfrac{1{,}67\cdot 10^{-27}\cdot 500\cdot 10^{3}}{1{,}6\cdot 10^{-19}\cdot 30\cdot 10^{-3}}R = 0{,}174\,\text{м}
  2. T = \dfrac{2\pi R}{v} = \dfrac{2\pi m}{qB}T = \dfrac{2\cdot 3{,}14\cdot 0{,}174}{500\cdot 10^{3}}T = 2{,}2\cdot 10^{-6}\,\text{с}

    The period does not depend on the particle speed.

AnswerR = 0{,}174\,\text{м}T = 2{,}2\cdot 10^{-6}\,\text{с}

Position in this set: 6 from 23

Problem #6

Magnetism: magnetic field forces

In a magnetic field with induction 4\,\text{мТл} a horizontal conductor of length 20\,\text{см} and mass 16\,\text{г}. The conductor is perpendicular to the induction vector. Determine the current.

Answer options

  1. 100\,\text{А}
  2. 400\,\text{А}
  3. 2000\,\text{А}
  4. 200\,\text{А}
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Given
B
4\cdot 10^{-3}\,\text{Тл}
\ell
0{,}2\,\text{м}
m
0{,}016\,\text{кг}
\alpha
90^{\circ}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

I

  1. F_A = mg

    Equilibrium: Ampere's force balances the force of gravity.

  2. I\ell B\sin\alpha = mg\ \Rightarrow\ I = \dfrac{mg}{\ell B\sin\alpha}I = \dfrac{0{,}016\cdot 10}{0{,}2\cdot 4\cdot 10^{-3}\cdot 1}I = 200\,\text{А}

AnswerI = 200\,\text{А}

Position in this set: 7 from 23

Problem #7

Magnetism: electromagnetic induction

Coil diameter 5\,\text{см} has 200 turns and is in a magnetic field parallel to the axis of the coil. What is the EMF if 0{,}3\,\text{с} induction changed from 80\,\text{мТл} up to 20\,\text{мТл}?

Answer options

  1. 0{,}08\,\text{В}
  2. 0{,}04\,\text{В}
  3. 0{,}16\,\text{В}
  4. 0{,}8\,\text{В}
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Given
d
5\,\text{см}0{,}05\,\text{м}
n
200
\Delta t
0{,}3\,\text{с}
B_1
80\cdot 10^{-3}\,\text{Тл}
B_2
20\cdot 10^{-3}\,\text{Тл}
Find

\varepsilon_i

  1. S = \dfrac{\pi d^{2}}{4}S = \dfrac{3{,}14\cdot 0{,}05^{2}}{4}S = 0{,}002\,\text{м}^{2}
  2. \left|\Delta\Phi\right| = \left|B_2 - B_1\right|S

    The field is parallel to the coil axis, so the cosine is equal to one.

  3. \varepsilon_i = \dfrac{n\left|\Delta\Phi\right|}{\Delta t}\varepsilon_i = \dfrac{200\cdot\left|20\cdot 10^{-3} - 80\cdot 10^{-3}\right|\cdot 0{,}002}{0{,}3}\varepsilon_i = 0{,}08\,\text{В}

Answer\varepsilon_i = 0{,}08\,\text{В}

Position in this set: 8 from 23

Problem #8

Magnetism: electromagnetic induction

When the current in the coil changes by inductance 0{,}2\,\text{Гн} for 0{,}1\,\text{с} self-induced emf occurred 3\,\text{В}. How much has the current changed?

Answer options

  1. 15\,\text{А}
  2. 1{,}5\,\text{А}
  3. 0{,}75\,\text{А}
  4. 3\,\text{А}
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Given
\Delta t
0{,}1\,\text{с}
L
0{,}2\,\text{Гн}
\varepsilon_{si}
3\,\text{В}
Find

\Delta I

  1. \varepsilon_{si} = \dfrac{L\left|\Delta I\right|}{\Delta t}\ \Rightarrow\ \left|\Delta I\right| = \dfrac{\varepsilon_{si}\,\Delta t}{L}\left|\Delta I\right| = \dfrac{3\cdot 0{,}1}{0{,}2}\Delta I = 1{,}5\,\text{А}

Answer\Delta I = 1{,}5\,\text{А}

Position in this set: 9 from 23

Problem #9

Magnetism: electromagnetic induction

In a coil with inductance 0{,}5\,\text{Гн} current flows 8\,\text{А}. What is the energy of the magnetic field? How to change the current so that the energy increases in 9 once?

Answer options

  1. \left(32\,\text{Дж};\ 48\,\text{А}\right)
  2. \left(160\,\text{Дж};\ 240\,\text{А}\right)
  3. \left(16\,\text{Дж};\ 24\,\text{А}\right)
  4. \left(8\,\text{Дж};\ 12\,\text{А}\right)
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Given
L
0{,}5\,\text{Гн}
I
8\,\text{А}
Find

W_{\text{м}}I'

  1. W_{\text{м}} = \dfrac{LI^{2}}{2}W_{\text{м}} = \dfrac{0{,}5\cdot 8^{2}}{2}W_{\text{м}} = 16\,\text{Дж}
  2. W \sim I^{2}\ \Rightarrow\ I \sim \sqrt{W}I' = 3I = 24\,\text{А}

    Energy grows as the square of the current, so to increase by 9 times the current must be tripled.

AnswerW_{\text{м}} = 16\,\text{Дж}I' = 24\,\text{А}

Position in this set: 10 from 23

Problem #10

Magnetism: electromagnetic induction

Closed resistance conductor 3\,\text{Ом} is in a magnetic field. What charge will pass through the conductor when the magnetic flux changes by 3\cdot10^{-4}\,\text{Вб}?

Answer options

  1. 5 \cdot 10^{-5}\,\text{Кл}
  2. 2 \cdot 10^{-4}\,\text{Кл}
  3. 0{,}001\,\text{Кл}
  4. 1 \cdot 10^{-4}\,\text{Кл}
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Given
R
3\,\text{Ом}
\Delta\Phi
3\cdot 10^{-4}\,\text{Вб}
Find

q

  1. \varepsilon_i = \dfrac{\left|\Delta\Phi\right|}{\Delta t},\quad I = \dfrac{\varepsilon_i}{R}

    The induction current is determined by Faraday's law and Ohm's law.

  2. q = I\,\Delta t = \dfrac{\left|\Delta\Phi\right|}{R}q = \dfrac{3\cdot 10^{-4}}{3}q = 1\cdot 10^{-4}\,\text{Кл}

    Time is reduced: charge depends only on changes in flux and resistance.

Answerq = 1\cdot 10^{-4}\,\text{Кл}

Position in this set: 11 from 23

Problem #11

Magnetism: magnetic field forces

An electron flies perpendicularly into a magnetic field with a speed 5\,\text{Мм/с} and moves in a circle with radius 1{,}1\,\text{см}. Determine the field induction and revolution period.

Answer options

  1. \left(0{,}0026\,\text{Тл};\ 1{,}38 \cdot 10^{-8}\,\text{с}\right)
  2. \left(0{,}0013\,\text{Тл};\ 6{,}9 \cdot 10^{-9}\,\text{с}\right)
  3. \left(0{,}0052\,\text{Тл};\ 2{,}76 \cdot 10^{-8}\,\text{с}\right)
  4. \left(0{,}026\,\text{Тл};\ 1{,}38 \cdot 10^{-7}\,\text{с}\right)
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Given
\alpha
90^{\circ}
v
5\,\tfrac{\text{Мм}}{\text{с}}5\cdot 10^{6}\,\text{м/с}
R
1{,}1\,\text{см}1{,}1\cdot 10^{-2}\,\text{м}
q
1{,}6\cdot 10^{-19}\,\text{Кл}
m
9{,}1\cdot 10^{-31}\,\text{кг}
Find

BT

  1. qvB = \dfrac{mv^{2}}{R}\ \Rightarrow\ B = \dfrac{mv}{qR}B = \dfrac{9{,}1\cdot 10^{-31}\cdot 5\cdot 10^{6}}{1{,}6\cdot 10^{-19}\cdot 1{,}1\cdot 10^{-2}}B = 2{,}6\cdot 10^{-3}\,\text{Тл}
  2. T = \dfrac{2\pi R}{v}T = \dfrac{2\cdot 3{,}14\cdot 1{,}1\cdot 10^{-2}}{5\cdot 10^{6}}T = 1{,}38\cdot 10^{-8}\,\text{с}

AnswerB = 2{,}6\cdot 10^{-3}\,\text{Тл}T = 1{,}38\cdot 10^{-8}\,\text{с}

Position in this set: 12 from 23

Problem #92

Magnetic field and forces

Conductor length 0{,}2 m with current 5 A is located perpendicular to the magnetic field 0{,}4 Tl. Find the Ampere force.

Answer options

  1. 0{,}4\,\text{Н}
  2. 0{,}8\,\text{Н}
  3. 0{,}2\,\text{Н}
  4. 1{,}2\,\text{Н}
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Given
l
0{,}2\,\text{м}
I
5\,\text{А}
B
0{,}4\,\text{Тл}
Find

F

  1. F=BIlF=0{,}4 \cdot 5 \cdot 0{,}2F=0{,}4\,\text{Н}

    The angle is 90°, the sine is equal to one.

Answer0{,}4\,\text{Н}

Position in this set: 13 from 23

Problem #93

Magnetic field and forces

Conductor length 0{,}4 m with current 3 A is located perpendicular to the magnetic field 0{,}5 Tl. Find the Ampere force.

Answer options

  1. 1{,}2\,\text{Н}
  2. 0{,}6\,\text{Н}
  3. 0{,}3\,\text{Н}
  4. 1{,}8\,\text{Н}
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Given
l
0{,}4\,\text{м}
I
3\,\text{А}
B
0{,}5\,\text{Тл}
Find

F

  1. F=BIlF=0{,}5 \cdot 3 \cdot 0{,}4F=0{,}6\,\text{Н}

    The angle is 90°, the sine is equal to one.

Answer0{,}6\,\text{Н}

Position in this set: 14 from 23

Problem #94

Magnetic field and forces

Horizontal conductor with mass 20 g and length 0{,}2 m is in a perpendicular field 0{,}5 Tl. What current is needed for the upward force of Ampere to balance the gravity? g = 10 m/s².

Answer options

  1. 4\,\text{А}
  2. 2\,\text{А}
  3. 1\,\text{А}
  4. 6\,\text{А}
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Given
m
20\,\text{г}
l
0{,}2\,\text{м}
B
0{,}5\,\text{Тл}
g
10\,\text{м/с}^{2}
Find

I

  1. F=mgF=\frac{20}{1000} \cdot 10F=0{,}2\,\text{Н}

    Convert the mass into kilograms and find the force of gravity.

  2. I=\frac{F}{Bl}I=\frac{0{,}2}{(0{,}5 \cdot 0{,}2)}I=2\,\text{А}

    At equilibrium BIl = mg.

Answer2\,\text{А}

Position in this set: 15 from 23

Problem #95

Magnetic field and forces

Horizontal conductor with mass 30 g and length 0{,}3 m is in a perpendicular field 0{,}5 Tl. What current is needed for the upward force of Ampere to balance the gravity? g = 10 m/s².

Answer options

  1. 4\,\text{А}
  2. 1\,\text{А}
  3. 2\,\text{А}
  4. 6\,\text{А}
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Given
m
30\,\text{г}
l
0{,}3\,\text{м}
B
0{,}5\,\text{Тл}
g
10\,\text{м/с}^{2}
Find

I

  1. F=mgF=\frac{30}{1000} \cdot 10F=0{,}3\,\text{Н}

    Convert the mass into kilograms and find the force of gravity.

  2. I=\frac{F}{Bl}I=\frac{0{,}3}{(0{,}5 \cdot 0{,}3)}I=2\,\text{А}

    At equilibrium BIl = mg.

Answer2\,\text{А}

Position in this set: 16 from 23

Problem #96

Magnetic field and forces

Particle with charge 0{,}000001 Cl and mass 1\cdot 10^{-9} kg moves perpendicular to the field 0{,}2 Tl with speed 100 m/s. Find the radius of the circle.

Answer options

  1. 1\,\text{м}
  2. 0{,}25\,\text{м}
  3. 0{,}5\,\text{м}
  4. 1{,}5\,\text{м}
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Given
q
0{,}000001\,\text{Кл}
m
1\cdot 10^{-9}\,\text{кг}
B
0{,}2\,\text{Тл}
v
100\,\text{м/с}
Find

r

  1. F=|q|vBF=0{,}000001 \cdot 100 \cdot 0{,}2F=0{,}00002\,\text{Н}

    Magnetic force is perpendicular to speed.

  2. a=\frac{F}{m}a=\frac{0{,}00002}{1\cdot 10^{-9}}a=20000\,\text{м/с}^{2}

    This is centripetal acceleration.

  3. r=\frac{v^2}{a}r=\frac{{100}^{2}}{20000}r=0{,}5\,\text{м}

    We obtain the radius from a = v²/r.

Answer0{,}5\,\text{м}

Position in this set: 17 from 23

Problem #97

Magnetic field and forces

Particle with charge 0{,}000002 Cl and mass 2\cdot 10^{-9} kg moves perpendicular to the field 0{,}4 Tl with speed 200 m/s. Find the radius of the circle.

Answer options

  1. 1\,\text{м}
  2. 0{,}25\,\text{м}
  3. 1{,}5\,\text{м}
  4. 0{,}5\,\text{м}
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Given
q
0{,}000002\,\text{Кл}
m
2\cdot 10^{-9}\,\text{кг}
B
0{,}4\,\text{Тл}
v
200\,\text{м/с}
Find

r

  1. F=|q|vBF=0{,}000002 \cdot 200 \cdot 0{,}4F=0{,}00016\,\text{Н}

    Magnetic force is perpendicular to speed.

  2. a=\frac{F}{m}a=\frac{0{,}00016}{2\cdot 10^{-9}}a=80000\,\text{м/с}^{2}

    This is centripetal acceleration.

  3. r=\frac{v^2}{a}r=\frac{{200}^{2}}{80000}r=0{,}5\,\text{м}

    We obtain the radius from a = v²/r.

Answer0{,}5\,\text{м}

Position in this set: 18 from 23

Problem #98

Magnetic flux and induction

Uniform magnetic field 0{,}2 T perpendicular to the plane of the frame with area 0{,}04 m². Find the magnetic flux.

Answer options

  1. 0{,}016\,\text{Вб}
  2. 0{,}008\,\text{Вб}
  3. 0{,}004\,\text{Вб}
  4. 0{,}024\,\text{Вб}
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Given
B
0{,}2\,\text{Тл}
S
0{,}04\,\text{м}^{2}
Find

\Phi

  1. \Phi=BS\Phi=0{,}2 \cdot 0{,}04\Phi=0{,}008\,\text{Вб}

    The normal to the frame is parallel to the field.

Answer0{,}008\,\text{Вб}

Position in this set: 19 from 23

Problem #99

Magnetic flux and induction

Uniform magnetic field 0{,}3 T perpendicular to the plane of the frame with area 0{,}06 m². Find the magnetic flux.

Answer options

  1. 0{,}036\,\text{Вб}
  2. 0{,}009\,\text{Вб}
  3. 0{,}018\,\text{Вб}
  4. 0{,}054\,\text{Вб}
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Given
B
0{,}3\,\text{Тл}
S
0{,}06\,\text{м}^{2}
Find

\Phi

  1. \Phi=BS\Phi=0{,}3 \cdot 0{,}06\Phi=0{,}018\,\text{Вб}

    The normal to the frame is parallel to the field.

Answer0{,}018\,\text{Вб}

Position in this set: 20 from 23

Problem #100

Magnetic flux and induction

The flux through each turn of the coil is from 100 turns decreased uniformly from 0{,}006 Wb to zero for 0{,}2 s. Find the module of the induced emf.

Answer options

  1. 6\,\text{В}
  2. 1{,}5\,\text{В}
  3. 3\,\text{В}
  4. 9\,\text{В}
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Given
N
100\,\text{витков}
F
0{,}006\,\text{Вб}
t
0{,}2\,\text{с}
Find

|\mathcal{E}|

  1. a=\frac{|\Delta\Phi|}{\Delta t}a=\frac{0{,}006}{0{,}2}a=0{,}03\,\text{Вб/с}

    Find the rate of change of flow after one turn.

  2. |\mathcal{E}|=Na|\mathcal{E}|=100 \cdot 0{,}03|\mathcal{E}|=3\,\text{В}

    The EMF of the turns adds up.

Answer3\,\text{В}

Position in this set: 21 from 23

Problem #101

Magnetic flux and induction

The flux through each turn of the coil is from 200 turns decreased uniformly from 0{,}004 Wb to zero for 0{,}4 s. Find the module of the induced emf.

Answer options

  1. 4\,\text{В}
  2. 1\,\text{В}
  3. 6\,\text{В}
  4. 2\,\text{В}
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Given
N
200\,\text{витков}
F
0{,}004\,\text{Вб}
t
0{,}4\,\text{с}
Find

|\mathcal{E}|

  1. a=\frac{|\Delta\Phi|}{\Delta t}a=\frac{0{,}004}{0{,}4}a=0{,}01\,\text{Вб/с}

    Find the rate of change of flow after one turn.

  2. |\mathcal{E}|=Na|\mathcal{E}|=200 \cdot 0{,}01|\mathcal{E}|=2\,\text{В}

    The EMF of the turns adds up.

Answer2\,\text{В}

Position in this set: 22 from 23

Problem #102

Magnetic flux and induction

A field perpendicular to a frame with an area 0{,}1 m², decreases evenly with 0{,}8 up to 0{,}2 Tl for 0{,}2 s. Frame resistance 3 Ohm. Find the magnitude of the induction current, neglect self-induction.

Answer options

  1. 0{,}2\,\text{А}
  2. 0{,}05\,\text{А}
  3. 0{,}3\,\text{А}
  4. 0{,}1\,\text{А}
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Given
S
0{,}1\,\text{м}^{2}
B_1
0{,}8\,\text{Тл}
B_2
0{,}2\,\text{Тл}
t
0{,}2\,\text{с}
R
3\,\text{Ом}
Find

I

  1. |\Delta\Phi|=(B_1-B_2)S|\Delta\Phi|=(0{,}8 - 0{,}2) \cdot 0{,}1|\Delta\Phi|=0{,}06\,\text{Вб}

    The frame area does not change.

  2. |\mathcal{E}|=\frac{|\Delta\Phi|}{t}|\mathcal{E}|=\frac{0{,}06}{0{,}2}|\mathcal{E}|=0{,}3\,\text{В}

    We use the law of induction.

  3. I=\frac{|\mathcal{E}|}{R}I=\frac{0{,}3}{3}I=0{,}1\,\text{А}

    The current module is determined by Ohm's law.

Answer0{,}1\,\text{А}

Position in this set: 23 from 23

Problem #103

Magnetic flux and induction

A field perpendicular to a frame with an area 0{,}2 m², decreases evenly with 0{,}9 up to 0{,}3 Tl for 0{,}3 s. Frame resistance 4 Ohm. Find the magnitude of the induction current, neglect self-induction.

Answer options

  1. 0{,}1\,\text{А}
  2. 0{,}2\,\text{А}
  3. 0{,}05\,\text{А}
  4. 0{,}3\,\text{А}
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Given
S
0{,}2\,\text{м}^{2}
B_1
0{,}9\,\text{Тл}
B_2
0{,}3\,\text{Тл}
t
0{,}3\,\text{с}
R
4\,\text{Ом}
Find

I

  1. |\Delta\Phi|=(B_1-B_2)S|\Delta\Phi|=(0{,}9 - 0{,}3) \cdot 0{,}2|\Delta\Phi|=0{,}12\,\text{Вб}

    The frame area does not change.

  2. |\mathcal{E}|=\frac{|\Delta\Phi|}{t}|\mathcal{E}|=\frac{0{,}12}{0{,}3}|\mathcal{E}|=0{,}4\,\text{В}

    We use the law of induction.

  3. I=\frac{|\mathcal{E}|}{R}I=\frac{0{,}4}{4}I=0{,}1\,\text{А}

    The current module is determined by Ohm's law.

Answer0{,}1\,\text{А}

The theory behind these problems

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Magnetic field and forcesMagnetic flux density, forces on currents and charges, and charged-particle motion.Magnetic flux and inductionFlux, electromagnetic induction, self-induction and field energy.

Formulas by topic

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InductanceL · ГнMagnetic flux densityB · ТлMagnetic fluxΦ · ВбVelocityv · м/сEnergyE, W · Дж