Slo BasednElectromagnetism Physics class 10 chapter 15 Conceptual Questions , Comprehensive Questions, Numerical Questions, and review questions and answers.
Slo Based Conceptual Questions Physics 10 Class Notes
Can an electron at rest be set into motion with a magnetic field?
No, the magnetic field only influences the direction of a moving charged body. At rest, only an electric field is associated with the charge. Hence, an electron at rest will not be influenced by a magnetic field.
Which is more likely to show deflection in compass needle, AC current, or DC current? Explain.
An AC current will more likely to show a deflection in the compass needle. The reason is that a magnetic field is associated with the alternating current. And a compass needle is used to detect a magnetic field.
On the other hand, no magnetic field is associated with the DC current. Hence, it does not show any deflection in the magnetic field.
A constant magnetic field is applied to a current-carrying conductor. What angle should the wire make with the field for the force to be (a) maximum, (b) minimum?
As we know:
FB = q (v × B) = q vB sinθ
sin 0 = 0 and sin 90 = 1
(a) maximum:
It can be seen from the above equations that the force will be maximum when the angle between the magnetic field and wire is 90o.
(b) minimum:
The force will be minimum when the angle between the magnetic field and wire is 0o.
Why does a compass needle point North?
A compass needle is designed in a way that in the absence of an external magnetic field, the compass needle normally points in the earth’s north direction.
The compass needle is made up of a permanent magnet. And naturally, the north pole is attracted towards the south pole. So it means that the needle head is made of the south magnetic pole so that it always direct towards the north pole.
How can a magnetic field be used to generate electric current?
Yes, electric current can be generated by a magnetic field using the phenomenon of electromagnetic induction.
When the magnetic field through a wire is changed, a current is induced in the wire, this phenomenon is known as electromagnetic induction.
What would happen if we use a slip ring to drive a DC motor?
If we use slip rings instead of split ring the direction of the current in the coil remains the same after a half cycle. Due to which the coil will rotate in the opposite direction. The coil would oscillate (wobble) back and forth.
The primary coil of a transformer is connected to a DC battery. Is there an emf induced in the secondary coil? Why?
No current will be induced in the secondary coil. A transformer works on the principle of mutual induction. A DC battery produces a continuous and non-varying current, which cannot produce a magnetic flux in the primary coil. Hence, no flux is induced in the secondary coil.
Comprehensive Questions Physics 10 Class Notes Chapter 15
Q.1) Describe an experiment to show that the steady current carrying wire produces a magnetic field around it. What is the direction of this magnetic field?
Answer:
When current is passed through a long wire with compass needles placed around it in a circle, the compass needles line up with the magnetic field in a circular pattern. This experiment shows the orientation of a magnetic field around a current-carrying wire is circular. Without any current in the wire all the compass needle will align with earth’s magnetic field, pointing towards the north.
As soon as the current is passed through the wire, the compass needles will redistribute to align with the magnetic field of the wire. The compass needle deflects in the opposite direction when the current is reversed. The direction of such a field is determined by the right-hand rule.
Right-Hand Rule I:
“Put your thumb in the direction of the current and curl your fingers around the wire, the curled fingers will show the direction of the magnetic field.”
This rule is for the direction of conventional current or flow of positive charges. For electronic current flow, the same rule is applied but with the left hand.
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Q.2) How the magnetic field of wire increases and resembles more like that from a permanent magnet if the wire is formed in a circular coil?
[Hint: Consider the magnetic field of a single coil and then a series of coils.]
Answer:
Magnetic Field due to a coil or solenoid:
When a wire is shaped into the loop or coil and steady current is applied through it. By applying the right-hand ode around a loop of wire carrying current, we see that the magnetic field around a loop of wire carrying current is in the same direction coming in and going out of the coil, forming north and a south pole like a bar magnet. Since the field lines cannot cross, the field lines are packed closer together inside therefore the field is stronger inside and weaker outside the coil.
To make a field having resemblance with the field of a bar magnet, a coil is shaped as a spiral (helix) called solenoid. When current is passed through a solenoid, a reasonably uniform magnetic field can be produced. This field is strong along the axis of solenoid and weaker outside. For a tightly wound solenoid, the field in the interior space is very uniform and strong. The field lines have a resemblance to those of a bar magnet, meaning that the solenoid effectively has north and south poles. To determine the direction we use right-hand rule II.
Right-hand rule II for a solenoid:
Curl the fingers in the direction of current around the coil or solenoid and the extended thumb will point in the direction of the north pole of a magnet. This rule is for the direction of conventional current or flow of positive charges. For electronic current flow, the same rule is applied but with the left hand.
Q.3) Explain the force on a current carrying wire in a magnetic field.
Answer:
Force on a current-carrying conductor in a Magnetic Field:
When a current-carrying wire is placed in a magnetic field, charges flowing through the wire interact with the external magnetic field and thus the wire experiences a force. Consider a wire connected to a battery and passing through a U-shaped magnet. It is observed that the wire experiences a force which is perpendicular to both the direction of the current I and the direction of the magnetic field B. If we reverse the direction of the current, the direction of force is also reversed.
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We see that this force is maximum when the current-carrying wire is placed perpendicular to the magnetic field and reduces in magnitude when the angle is changed and vanishes when the current-carrying wire and magnetic field are Parallel, showing a sine of angle θ variations.
With this experiment we can find that this force is directly proportional to the current flowing though the wire I, the magnetic field B, length L of the wire in the Magnetic field and sine of angle B between current and magnetic field. Thus. mathematically the force is
FB ∝ BIL sinθ
FB = k BIL sinθ
For SI units K = 1, therefore the equation that describes the force experienced by a current-carrying conductor in a uniform external magnetic field then becomes
FB = BIL sinθ
Where F is the magnetic force in newtons (N), B is the magnetic field strength in testa (T), I is the current in the conductor in amperes (A), L is the length of the conductor in the magnetic field in metres (m), and θ is the angle between the conductor and the magnetic field.
The direction of force is determined by Fleming’s left-hand rule which is stated as:
“Stretch the thumb, forefinger and the middle finger of the left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the middle finger in the direction of the current, then the thumb would indicate the direction of the force acting on the conductor”
Q.4) Explain the torque on current Carrying coil in a Magnetic field.
Answer:
Turning Effect on a current-carrying coil in a magnetic field:
When a current-carrying coil or loop is placed in a uniform magnetic field it experiences a net force, this force can exert torque on the coil or loop of wire.
Consider a rectangular coil having four (1, 2, 3 and 4) sides of length a and b carrying a current I in the presence of a uniform magnetic field B directed parallel to the plane of the loop.
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No magnetic forces act on sides 1 and 3 because these wires are parallel to the field θ = 0o; the length of sides 1 and 3 is b, therefore magnetic force F1 and F3 is
F1 = F3 = Blb sin0° since sin0° = 0
Therefore F1 = F3 = 0
However, maximum magnetic forces act on sides 2 and 4, because these sides are perpendicular to the field θ = 90o; the length of sides 2 and 4 is a the magnetic force is
F2 = F4 = BILsin90o since sin90° = 1
Therefore F2 = F4 = Bla
If we view the loop from side 3 and sight along sides 2 and 4, we see the view shown in figure (b). The two magnetic forces F2 and F4 point in opposite directions such that they form a couple. If the loop is pivoted so that it can rotate about point 0, these two forces produce a torque that rotates the loop (in this case clockwise).
Q.5) Explain the working of DC motor.
Answer:
D.C Motor:
It is an electric motor which converts provided electrical energy into mechanical energy.
Systematic Diagram
Construction:
A D.C motor consists of a rectangular coil, say abcd mounted on a spindle such that it can move easily between the two poles of a permanent magnet as shown in the diagram. On the spindle, a copper ring is fitted which is divided into two halves say s1 and s2. These rings are tightly held by two carbon brushes X and Y with the help of a spring. One end of the rectangular coil is fixed in s1 and other in s2.
Working:
Let when we switch on the current then at that time, coil abcd be in the horizontal position. The coil will start to rotate as we connect the carbon brushes to the battery. The current will start flowing in a different direction as shown in figures by arrows. According to Fleming’s left-hand rule, the force will act in an upward direction on side ab and on cd the force will act in the downward direction. As a result of these two forces of equal magnitude but opposite direction, a torque will generate and the coil will start to rotate. Rotation of clock will be in the clockwise direction.
Q.6) Describe the phenomena of electromagnetic induction. List the factors effecting electromagnetic induction.
Answer:
Electromagnetic Induction:
Consider a loop of conducting wire connected to a galvanometer. Now if we bring a magnet and move it towards the coil of wire then there will be some deflection in the galvanometer and if we move the magnet away from the coil then galvanometer will show deflection again but in opposite direction to first one. There will also be deflection in needle if we hold a magnet stationary close to the coil.
These deflections of the needle of galvanometer mean that there is some flow of current through the coil when we bring the magnet closer or away from the coil. The relative motion towards or away from the coil results into change in magnetic flux through the coil and therefore this change in flux results into generation of current in the coil of wire.
Factors affecting electromagnetic induction:
- It is directly proportional to the number of loops.
- It depends on the speed at which the conductor moves through the magnetic field.
- It is directly proportional to the length of the conductor.
- It is directly proportional to the magnetic field of the magnets.
Q.7) Explain the direction of induced emf and show its relation to conservation of energy.
Answer:
Direction of induced emf and conservation of energy:
Whenever there is a change of magnetic flux, there will be induced emf and current will flow in such a direction so as to oppose the cause producing it.
The induced current around a loop produces its own magnetic field. This field may be weak compared with the external magnetic field. It cannot prevent the magnetic flux through the loop from changing, but its direction is always such that it ” tries ” to prevent the flux from changing.
Consider pushing the bar magnet’s North pole into the loop, it causes the magnetic field to increase in the upward direction. To oppose the change, the loop itself needs to generate the downward-pointing magnetic field (acting as a magnet with its North Pole at the bottom which repels the North pole of the bar magnet moving towards it), as shown in figure (a).
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The induced magnetic field at the centre of the loop will point downward (using right-hand rule II) only if the current is clockwise (cw) as seen from above. Now suppose the north pole of the bar magnet is pulled away from the loop, as shown in figure (b). There is an upward magnetic field through the loop, but the magnetic field is decreasing as the magnet is moving away. Thus, the induced magnetic field of the loop opposes this decrease, therefore, the induced field needs to point in the upward direction, the induced current is counter clockwise (ccw) as seen from above.
Similar effects can also be observed with the south pole of a magnet pushed in and out of the loop figure (c,d). It is seen that the induced current sets up a magnetic field of its own. From the above discussion, it is clear that the motion of magnet is always opposed by the magnetic field generated from induced current. The mechanical energy spent in overcoming this opposition is converted into electrical energy.
If the induced current were in the opposite direction, the magnetic force would have accelerated it with no external energy source, even though electric energy would have been dissipated in the circuit. This would have been a clear violation of the law of conservation of energy, which doesn’t happen in nature.
Q.8) Sketch and describe the construction and working of AC generator.
Answer:
AC Generator:
AC generator works on the principle of electromagnetic induction. Using this phenomenon, an AC generator converts mechanical energy into electrical energy.
Schematic Diagram:
Explanation and working:
An AC generator consists of three parts as follows:
- Magnetic field.
- An armature.
- Slip Rings.
- Brushes.
In an AC generator, a rectangular coil is placed between two pieces of a very large permanent magnet. The two of these coils are connected to two circular rings which are further connected to two carbon brushes permanently as shown in the figure. When we supply mechanical energy to this coil then it will rotate and as a result, an induced emf will generate across its ends. As during the rotation of the coil, it will make different angles and this produced emf will be alternating emf and therefore the generator is called alternating current generator. And coil of the alternating generator is known as an armature.
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Applications:
EMF which is generated from an alternating current generator is used for running different machines which in turn provide light to our house.
Q.9) Describe the phenomena of mutual induction.
Answer:
Mutual Induction:
“The phenomenon in which emf is induced in a secondary coil or circuit due to the change in current in the primary coil or circuit is called mutual induction.”
Systematic Diagram:
Explanation
It consists of two coil which are placed side by side to each other. The first coil which is connected to a battery is known as the primary coil, while the coil which does not have any emf source is known as the secondary coil.
When the circuit is disconnected i.e. when the switch is open then there will not be any current and therefore no magnetic lines will pass through these coils. However when the switch is closed then there will be a current flow through the first coil and hence there will be a magnetic field. Some of these magnetic lines of force will also pass through the secondary coil as well. As the current through the first coil will change then the magnetic field through the second secondary coil will also change.
According to Faraday’s law of electromagnetic induction, due to this varying number of magnetic lines of force in the secondary coil, an induced current will be produced momentarily and hence the induced emf as well. This induced emf depends on the following factors:
- The number of turns of two coils.
- Distance between two coils.
- The cross-sectional area of each coil.
Q.10) What are transformers? On what Principle it works? Also describe the purpose of transformers in AC circuits.
Answer:
Transformer:
A transformer is a very important application of electromagnetic induction.
“A transformer is an electronic device which converts a low voltage input to a high voltage output and vice verca”Schematic Diagram:
Construction:
It consists of two coils. Each coil is wounded on an iron core as shown in the above diagram. One of these coils is connected to some alternating current or emf source and is known as the primary coil. Another coil is called a secondary coil.
Working:
When we switch on the current the alternating current will start to flow through the primary coil and process of mutual induction will take place. As the core of the wounded wire is of iron, therefore, this iron core and the secondary coil will make the resulting field lines linkage to be strong. As due to changing magnetic field, the magnetic flux will alter through the iron core and hence will result in the induction of an alternating current in the second coil.
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Mathematical Expression:
Transformer equation is given as,
Here Vs and VP are the induced emf in secondary and primary coils, respectively. Similarly, NS and NP are the number of turns of secondary and primary coils, respectively.
There are two types of transformers as follows:
Step-up Transformer:
The transformer which converts low electric voltage to high voltage is known as a step-up transformer. For a step-up transformer induced emf in the secondary coil will be greater than induced emf in the primary coil and also the number of turns in the secondary coil will be greater than the number of turns in the primary coil i.e.
VS > VP and NS > NP
Step-Down Transformer:
The transformer which converts high applied voltage to low voltage is known as a step-down transformer. For a step-down transformer induced emf in the secondary coil will be less than induced emf in the primary coil and also the number of turns in the secondary coil will be less than the number of turns in primary coil i.e.
VS < VP and NS < NP
Uses:
A transformer can be used for the transmission and distribution of the electric power from the grating plant to some distant industrial and other uses.
Numerical Questions Physics Chapter No 15 pdf
Q.1) A 1.5 m long wire carries a current of 5 A, at right angle to a uniform magnetic field of 0.04 T. Determine the force exerted on the wire.
Answer:
Given Data:
Length of Wire = L = 1.5 m
Current through wire = I = 5 A
Angle between wire and magnetic field = θ = 90o
Force of wire = Fm = 0.04 N
To Find:
Magnetic Induction = B = ?
Solution:
As magnetic force is given by,
Fm = BILSinθ ⇒ B = Fm/ILSinθ
Putting values, we get
Q.2) A wire carrying a direct current of 10.0 A is suspended 5.0 m east between a house and a garage perpendicular to the Earth’s magnetic field of 5.0 x 10-5 T. What is the magnitude of the force that acts on the conductor?
Answer:
Given Data:
Length of Wire = L = 5 m = 5 m
Current through wire = I = 10 A
Angle between wire and magnetic field = θ = 90o
Magnetic Induction = B = 5.0 × 10-5 T
To Find:
Force of wire = Fm = ?
Solution:
As magnetic force is given by,
Fm = BILSinθ
Putting values, we get
Fm = 5.0 × 10-5 × 10 × 5 × Sin900
Fm = 0.0025 N
Fm = 2.5 × 10-3
Fm = 2.5 mT
Q.3) A 10 cm wire at 30° to uniform magnetic field of 0.06 T is exerted by a force 0.024 N. What is the current flowing through the wire?
Answer:
Given Data:
Length of Wire = L = 10 cm = 0.1 m
Current through wire = I = 4 A
Angle between wire and magnetic field = θ = 30o
Magnetic Induction = B = 0.06T
Fm = 0.024 N
To Find:
Current through wire = I =?
Solution:
As magnetic force is given by,
Fm = BILSinθ
Putting values, we get
I = Fm/BL Sinθ
I = 0.024/ (0.06 × 0.1 × Sin 300)
I = 8 A
Q.4) If the current through the primary coil changes from -5 A to +5 A in 0.05 s, Such that the induced emf is 2.8 V. What is the mutual inductance?
Answer:
Given:
Initial current, Ii = -5 A
Final current, If = 5 A
Time taken, Δt = 0.05s
Emf induced, εs = 2.8 V
To find:
Mutual Inductance, M =?
Calculation:
εs = -M × ΔI/Δt
M = -εs × Δt/ΔI
M = -εs × Δt/(If – Ii)
M = -2.8 × 0.05/(5 – (-5))
M = -2.8 × 0.05/(5 + 5)
M = -2.8 × 0.05/10
M = – 14 × 10-3
M = – 14 mH
Q.5) A transformer connected to a 120-V AC line is to supply 9600 V for a neon sign. (a) What is the ratio of secondary to primary turns of the transformer? (b) If the transformer consisted of 275 primary windings, how many secondary windings would there be?
Answer:
Given Data:
Voltage of secondary coil = VS = 9600 V
Voltage of primary coil = VP = 120 V
Number of turns in primary coil = NP = 275
To Find:
Ratio of secondary to primary turns, Ns/Np=?
Number of turns in secondary coil = NS = ?
Solution:
(a) As the transformer equation is given by,
Putting values, we get the turn ratio
(b) Using the above equation and putting values
Q.6) How many turns would you want in the secondary coil of a transformer having 400 turns in the primary, if it were to reduce the voltage from 220 V AC to 3.0 V AC?
Answer:
Given Data:
Voltage of secondary coil = VS = 3 V
Voltage of primary coil = VP = 220 V
Number of turns in primary coil = NP = 400
To Find:
Number of turns in secondary coil = NS = ?
Solution:
As the transformer equation is given by,
Putting values, we get
Q.7) A transformer steps down a main supply of 220 V AC to operate a 12 V AC lamp. Calculate the turns ratio of the windings
Answer:
Given Data:
Voltage of secondary coil = VS = 12V
Voltage of primary coil = VP = 220V
To Find:
Turns ratio = NS/NP = ?
Solution:
As the transformer equation is given by,
Putting values, we get
