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Physics [2 Mark Question Answer]

Magnetic Effect of Current · ICSE STD 10 (ICSE - English Medium). 26 questions.

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[2 Mark Question Answer]

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Question 12 Marks
A transformer lowers e.m.f. from 220 V to 15 V. If the number of turns in primary are 3520, how many turns are in the secondary coil?
Answer
We know that $\frac{ E _{ s }}{ E _{ p }}=\frac{ N _{ s }}{ N _{ p }}$
or $\frac{10}{220}=\frac{ N _{ s }}{3250}$
$
N _{ s }=\frac{3520 \times 15}{220}=240
$
Number of turns in secondary are $=240$.
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Question 22 Marks
A transformer lowers e.m.f. from 220 V to 15 V. If 400 W power is given in primary, calculate (i) the current in primary coil and (ii) the current in secondary coil.
Answer
(i) We know that Power $=I_p \times E_p$
or $400=I_p \times 220$
$
\therefore I_p=1.8 A
$
(ii) Using $I_p \times E_p=I_s \times E_s$
$
\therefore I _{ s }=\frac{ I _{ p } \times E _{ p }}{ E _{ s }}
$
$
I _{ s }=\frac{1.8 \times 220}{15}
$
$
=26.4 A \text {. }
$
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Question 32 Marks
A transformer lowers e.m.f. 220 V to 12 volts. If the number of turns in primary are 8800, how many turns are in secondary coil?
Answer
$
\begin{aligned}
& \text { Using } \frac{ E _{ p }}{ E _{ s }}=\frac{ N _{ p }}{ N _{ s }} \\
& \therefore \frac{200 V }{12 V }=\frac{8800}{ N _{ s }} \\
& \therefore N _{ s }=\frac{8800 \times 12}{220}=40 \times 12=480
\end{aligned}
$
$\therefore$ Number of turns in secondary coil $=480$.
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Question 42 Marks
Applying e.m. f to primary coil is 210 V. If the number of turns in primary coil is 200 turns and that of in secondary coil is 20 turns, then find out the output voltage. Name the type of transformer.
Answer
$
\begin{aligned}
& \text { Using } \frac{ E _{ p }}{ E _{ s }}=\frac{ N _{ p }}{ N _{ s }} \\
& \frac{210}{ E _{ s }}=\frac{200}{20} \\
& \text { or } E _{ s }=\frac{210 \times 20}{200}
\end{aligned}
$
Output voltage $=21 V$
It is step down transformer.
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Question 52 Marks
State a law, which determines the direction of magnetic field around a current carrying wire.
Answer
Right hand thumb rule determines the direction of magnetic field around a current carrying wire.
It states that if we hold the current carrying conductor in right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic fields lines.
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Question 62 Marks
A straight wire is held vertical and current is passed in it in the upward direction. A compass needle is kept near the wire in north of it. How is the needle deflected? What is the change in the deflection of the needle if the current in wire is reversed?
Answer
When the current in the wire is upwards, the needle will deflect with its north pole towards west.
If the current in the wire is reversed, the north pole of the needle will deflect towards east.
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Question 72 Marks
Describe briefly one way of producing an induced current in a coil. State one factor that determines the magnitude of induced e.m.f.
Answer
To produce an induced current in a coil, a magnet is moved along the axis of coil. The magnitude of induced e.mi. depends on the strength of magnet.
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Question 82 Marks
Can a transformer be used with direct current source? Give reason.
Answer
No, a transformer cannot be used with direct current source. The reason is that with direct current source, there will not be any change in the magnetic flux linked with the primary coil, so no e.m.f. will be induced in the secondary coil.
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Question 92 Marks
How are the e.m.f in the primary and secondary coils of a transformer related with the number of turns in these coils?
Answer
The relation is following:
$
\frac{\text { E.M.f. in secondary coil }}{\text { E.m.f. in primary coil }}=\frac{\text { Number of turns in primary coil }}{\text { Number of turns in secondary coil }}
$
S stands for secondary and $P$ stands for primary.
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Question 102 Marks
Suggest two ways in an a.c. generator to produce a higher e.m.f.
Answer
Two ways in an a.c generator to produce a higher e.m.f. are:
(1) By increasing the speed of rotation of the coil.
(2) By increasing the number of turns of coil.
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Question 112 Marks
State the principle of a D. C. motor.
Answer
It is based on the principle that when an electric current is passed through a conductor placed normally in a magnetic field, a force acts on the conductor as a result of which the conductor begins to move. The direction of the force is obtained with the help of Fleming’s left hand rule.
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Question 122 Marks
State the factors that can increase the speed of rotation of the coil in a D.C. motor.
Answer
The speed of rotation of the coil in a D.C. motor can be increased by:
(i) Increasing the strength of the current, (ii) Having more turns in the coil,
(iii) Increasing the area of the coil, (iv) By increasing the strength of the magnetic field.
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Question 132 Marks
What will happen to a compass needle when the compass is placed below a wire and a current is made to flow through the wire? Give a reason to justify your answer.
Answer
A current carrying conductor produces a magnetic field around it and the magnetic needle in this magnetic field experience a torque due to which it deflects to align itself in the direction of magnetic field.
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Question 142 Marks
What is an electric motor? State its principle.
Answer
An electric motor is a device which converts the electrical energy into the mechanical energy.
The working of an electric motor is based on the principle that a current carrying coil placed in a magnetic field with its plane parallel to the field, Experiences a deflecting couple.
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Question 152 Marks
State two advantages of an electromagnet over a permanent magnet.
Answer
1. An electromagnet can produce a strong magnetic field.
2. The strength of the magnetic field of an electromagnet can easily be changed by changing the current in its solenoid.
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Question 162 Marks
List some of the practical applications of an electromagnet.
Answer
Electromagnets are used to lift heavy loads of scrap iron etc., to separate magnetic substances from non-magnetic ones and in devices like the electric bell, telegraphs, telephones, loud speakers, etc.
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Question 172 Marks
Name two factors on which the magnitude of an induced e.m.f. in the
secondary coil depends.
Answer
The magnitude of e.m.f. induced in the secondary coil depends on the following two factors:
  1. The ratio of the number of turns in the secondary coil to the number of turns in the primary coil (i.e., turns ratio), and
  2. The magnitude of e.m.f applied in the primary coil.
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Question 182 Marks
Why is soft iron generally used as the core of the electromagnet?
Answer
(i) Soft iron has less retentivity so it acquires the magnetic properties only when the current flows through the coil wound on it and loses the magnetic properties as the current is switched off.
(ii) The soft iron intensifies the magnetic field of the electromagnet because of its high permeability.
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Question 192 Marks
State two ways by which the magnetic field due to a solenoid can be made stronger.
Answer
The magnetic field due to a solenoid can be made stronger by using:
(i) By increasing the number of turns of winding in the solenoid.
(ii) By increasing the current through the solenoid.
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Question 202 Marks
A straight current-carrying conductor is kept in a magnetic field. It Experiences a force. State three factors on which the magnitude of force Experienced by the conductor depends. Under what condition will the conductor Experience no force?
Answer
The three conditions are: (i) Current in the conductor, (ii) Length of conductor, (iii) Strength of i magnetic field. The conductor will Experience no force if the direction of current in the conductor is parallel to the direction of magnetic field.
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Question 212 Marks
Why soft iron is preferred to be used as the core of the electromagnet of an electric bell?
Answer
Soft iron is chosen to make the core of the electromagnet of an electric bell on account of the following two reasons:
(i) On account of high magnetic permeability of soft iron, it can be easily magnetized and demagnetized. So it makes a strong magnet when current flows through the solenoid but it is almost demagnetized when no current flows through the solenoid. This fact helps in the working of the electric bell.
(ii) The soft iron intensifies the magnetic field of the solenoid and so it helps in making a strong magnet.
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Question 222 Marks
Name two factors on which the strength of magnetic field of an electromagnet depends and state how does it depend on the factors stated by you.
Answer
The strength of magnetic field of an electromagnet depends on:

(i) Number of turns: The strength of magnetic increases on increasing the number of turns of winding in the solenoid.
(ii) Current: The strength of magnetic field increases on increasing the current through the solenoid.

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Question 232 Marks
(i) A straight wire conductor passes vertically through a piece of cardboard sprinkled with iron filings. Copy the diagram and show the setting of iron filings when a current is passed through the wire in the upward direction and the cardboard is tapped gently. Draw arrows to represent the direction of the magnetic field lines.

(ii) Name the law which helped you to find the direction of the magnetic field lines.
Answer
(i)

(ii) Right hand thumb rule helped to find the direction of magnetic field lines.
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Question 252 Marks
You have been provided with a solenoid AB.

(i) What is the polarity at end A?
(ii) Give one advantage of an electromagnet over a permanent magnet.
Answer
(i) North pole.
(ii) The electromagnet is a very strong magnet as compared to a permanent magnet.
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Question 262 Marks
Draw a labelled diagram to make an electromagnet from a soft iron bar. Mark the polarity at its ends in your diagram. What precaution would you observe while making it?
Answer
The labelled diagram is shown in figure.=

Precaution: The source of current must be the D. C. battery.
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