Electromagnetic Induction Test

Result

Electromagnetic Induction Test - 22

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Complete the following statement. For electromagnetic induction to occur:

A
a magnet must be at rest within a coil of wire

B
a coil of wire must be at rest relative to the magnet.

C
a magnet must move through a coil of wire.

D
a magnet and a coil must have the same velocity.

E
a magnet must be pointing north.

Solution

Current is induced in a coil when flux through the coil changes. The flux through the coil changes when a magnet moves relative to the coil and coil is at rest relative to the magnet.
Hence the correct answer is option B.
Question 2
1 / -0

A helicopter rises vertically with a speed of 100 m/s. If helicopter has length 10 m and horizontal component of earth's magnetic field is $$5\times 10^{-3} Wb/m^2$$, then the induced emf between the tip of nose and tail of helicopter is:

A
50 V

B
0.5 V

C
5 V

D
25 V

Solution

In case of motional emf, the motion of the conductor in the field exerts a force on the free charge in the conductor, so that one end of the conductor becomes positive, while the other negative resulting in a potential difference across its ends due to which a non-conservative electric field is set up in the conductor. In steady state the magnetic force on the free charge is balanced by the electric force due to induced field.
$$qE = qvB$$
or $$q\left(\frac{V}{l}\right)=qvB$$
ie, $$V=Bvl$$
So, the induced emf between tip of nose and tail of helicopter is given by
$$e=Bvl$$
$$=5\times 10^{-3}\times 10\times 100$$
$$=5V.$$
Question 3
1 / -0

If circular coil with $$N_{1}$$ turns is changed in to a coil of $$N_{2}$$ turns. What will be the ratio of self inductances in both cases.

A
$$\dfrac {N_{1}}{N_{2}}$$

B
$$\dfrac {N_{2}}{N_{1}}$$

C
$$\dfrac {N_{1}^{2}}{N_{2}^{2}}$$

D
$$\sqrt {\dfrac {N_{1}}{N_{2}}}$$

Solution

If circular coil with N1 turns is changed in to a coil of N2 turns.

$$L=\dfrac{Nd \phi}{dt} $$
$$L_1=\dfrac{N_1d \phi}{dt} $$
$$L_2=\dfrac{N_2d \phi}{dt} $$
L is in Henries
N is the Number of Turns
Φ is the Magnetic Flux
Ι is in Amperes
$$\dfrac{L_1}{L_2}=\dfrac{N_1}{N_2} $$
Question 4
1 / -0

If coil is placed perpendicular to field lines then number of lines passing through coil are :

A
minimum

B
maximum

C
zero

D
may be max. or min.

Solution

The magnitude of the induced emf in a circuit is proportional to the time rate of magnetic flux linked with the circuit.

Consider a rectangular loop which rotates in a uniform magnetic field, around an axis in the plane of the loop and passing through the centres of opposite sides.

If the area of the loop is A and the field strength is B and the normal to the loop makes an angle phi with the field direction at any instant, the flux through the loop at any time is B A cos phi. The flux is a maximum when the loop is normal to the field $$(\phi =0)$$ and zero when the loop is aligned parallel to the field $$(\phi =90).$$

If the loop rotates with angular velocity w, then we can substitute phi = wt, and we can then write flux at any instant $$= A B cos wt.$$

The induced emf is $$E =\dfrac{ - d (A B cos wt)}{dt}$$

$$E= A B w sin wt.$$

The answer to your question follows from that last expression. The induced emf is a sinusoidal function. It will be a minimum (that is zero), when phi = wt = 0, when the plane of the coil is normal to the field, and it will be maximum when $$\phi = wt = 90$$, or when the plane of the loop is parallel to the field
Question 5
1 / -0

Alternative current generator is basically based upon :

A
Amperes law

B
Lenz's law

C
Faradays law

D
coulombs law

Solution

Alternative current generator is based on Faraday's Law.
Faraday's Law states that when the magnetic flux linking a circuit changes,an emf is induced in the circuit proportional to the rate of change of flux linkage.
Question 6
1 / -0

Flux density under trailing pole tips in case of generator will :

A
increase

B
decrease

C
either increase or decrease

D
none of the above

Solution

LPT=leading pole tip
TPT=tailling pole top
Considering armature reaction.
where GNA=Geometrical axis
MNA=Magnetic neutral axis
here clearly see that TPT side flux increases due to armature in case of generator.
Question 7
1 / -0

Alternating current is flowing in inductance L and resistance R. The frequency of source is $$\displaystyle\frac{\omega}{2\pi}$$. Which of the following statement is correct.

A
For low frequency the limiting value of impedance is L

B
For high frequency the limiting value of impedance is $$L\omega$$

C
For high frequency the limiting value of impedance is R

D
For low frequency the limiting value of impedance is $$L\omega$$

Solution

$$\begin{array}{l} \, \, As\, frequency\, approaches\, zero\, or\, DC,\, the\, inducators\, reac\tan ce\, would\, decrease\, tozero\, , \\ acting\, like\, a\, short\, circuit.\, this\, means\, inductive\, reac\tan ce\, is\, proportional\, to\, fequency \\ \, \, \, \, \, \, \, \, \, \, \, \, \, so\, ,\, for\, low\, frequency\, the\, { { limimiting } }\, \, value\, of\, impedance\, is\, L,\, and\, \, alternating\, \\ current\, is\, flowing\, in\, inductance\, L\, and\, resistance\, R.\, \, The\, frequency\, of\, source\, is\, \frac { \omega }{ { 2\pi } } . \\ so\, the\, correct\, option\, is\, A. \end{array}$$
Question 8
1 / -0

In a transformer, coefficient of mutual inductance between primary and secondary coil is 0.2 H. When current changes by 5 Na in the primary, then: the induced era in the secondary will be

A
0.5 V

B
1 V

C
1.5 V

D
2.0 V

Solution

$$e = M\dfrac{di}{dt} = 0.2\times5 = 1 V$$
Question 9
1 / -0

The dimensions of self-inductance L are :

A
$$[ML^2T^{-2}A^{-2}]$$

B
$$[ML^2T^{-1}A^{-2}]$$

C
$$[ML^2T^{-1}A^{-1}]$$

D
$$[ML^{-2}T^{-2}A^{-2}]$$

Solution

Self-inductance L $$= \dfrac {e}{\Delta i/\Delta t}=\dfrac{e\Delta t}{\Delta i}$$
$$\therefore $$ Unit of L$$=\dfrac{Volt-second}{Ampere}$$
$$=\dfrac{(Joule/Coulomb)second}{Ampere}$$
$$=\dfrac{Newton-metre-second}{Coulomb-ampere}$$
$$\Rightarrow $$ Unit of L$$= \dfrac{Newton-metere}{Ampere^2}$$
$$\therefore [L]=\dfrac{[MLT^2][L]}{[A^2]}=[ML^2T^{-2}A^{-2}]$$
Question 10
1 / -0

A coil carrying electric current is placed in uniform magnetic field, then :

A
Torque is formed

B
emf is induced

C
Both (a) and (b) are correct

D
None of the above

Solution

As magnetic field is uniform so its rate of change with time is zero. By Faraday's law of induction, no emf is induced.

But current carrying coil experience torque inside magnetic field, which is cross product of magnetic moment of the coil and the magnetic field.
$$\vec{\tau}=\vec{M}\times\vec{B}$$

Option A is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electromagnetic Induction Test - 22

Result

Electromagnetic Induction Test - 22

Score

-

Rank

-

SHARING IS CARING

Question 1

a magnet must be at rest within a coil of wire

a coil of wire must be at rest relative to the magnet.

a magnet must move through a coil of wire.

a magnet and a coil must have the same velocity.

a magnet must be pointing north.

Solution

Question 2

50 V

0.5 V

5 V

25 V

Solution

Question 3

$$\dfrac {N_{1}}{N_{2}}$$

$$\dfrac {N_{2}}{N_{1}}$$

$$\dfrac {N_{1}^{2}}{N_{2}^{2}}$$

$$\sqrt {\dfrac {N_{1}}{N_{2}}}$$

Solution

Question 4

minimum

maximum

zero

may be max. or min.

Solution

Question 5

Amperes law

Lenz's law

Faradays law

coulombs law

Solution

Question 6

increase

decrease

either increase or decrease

none of the above

Solution

Question 7

For low frequency the limiting value of impedance is L

For high frequency the limiting value of impedance is $$L\omega$$

For high frequency the limiting value of impedance is R

For low frequency the limiting value of impedance is $$L\omega$$

Solution

Question 8

0.5 V

1 V

1.5 V

2.0 V

Solution

Question 9

$$[ML^2T^{-2}A^{-2}]$$

$$[ML^2T^{-1}A^{-2}]$$

$$[ML^2T^{-1}A^{-1}]$$

$$[ML^{-2}T^{-2}A^{-2}]$$

Solution

Question 10

Torque is formed

emf is induced

Both (a) and (b) are correct

None of the above

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.