NCERT Exemplar Solutions Class 10 Science Magnetic Effects of Electric Current — Multiple Choice Questions Q1–12
NCERT Exemplar Problems, Class 10 Science — Magnetic Effects of Electric Current, Multiple Choice Questions. All 12 questions solved.
This unit is about the three-way connection between electricity and magnetism, and its twelve multiple-choice questions divide along those three lines.
- Current produces a field. A straight wire gives concentric circles, a loop gives a field like a short bar magnet, and a solenoid gives a field indistinguishable from a bar magnet’s. The right-hand thumb rule gives the direction.
- A field exerts a force on a current. That is what drives an electric motor, and Fleming’s left-hand rule gives the direction.
- A changing field induces a current. That is electromagnetic induction, the basis of the generator, and Fleming’s right-hand rule gives the direction.
Keeping the three rules apart is worth doing once, since Question 8 tests it directly:
| Rule | Gives the direction of | Used for |
|---|---|---|
| Right-hand thumb | the magnetic field round a current | a straight conductor |
| Fleming’s left hand | the force on a current in a field | the motor |
| Fleming’s right hand | the induced current | the generator |
Key insight. Field lines carry two pieces of information at once: their direction is the direction of the field, and their spacing is its strength — closer together means stronger. Nearly half the questions here are answered by reading one or other of those off a diagram, and Question 1 is about the one thing spacing does not mean.
Choose the correct answer from the given four options in the following questions:
Question 1
Choose the incorrect statement from the following regarding magnetic lines of field
Solution. Three of these are the standard properties of field lines; one is a misstatement of what parallel lines mean.
(a) is the definition of field direction — it is how the direction is fixed experimentally, by seeing where a compass needle’s north pole points. True.
(b) is true. Field lines are closed curves: outside a magnet they run from N to S, and inside the magnet they continue from S back to N, so each line closes on itself. There are no free magnetic poles for them to start or stop at.
(d) is true, and is the standard convention: closer lines mean a stronger field.
(c) is false, and it is the answer. Parallel, equidistant field lines do not mean zero field — they mean a uniform field, one of the same strength and direction everywhere. Zero field would be represented by no lines at all, not by evenly spaced ones.
The distinction matters because uniform fields are common and useful: the field inside a long solenoid is uniform, which is Question 10, and a uniform field is what Question 5 places an electron and a proton in.
(c) — parallel, equidistant field lines represent a uniform field, not a zero one. Zero field is shown by the absence of lines.
Question 2
If the key in the arrangement (Figure 13.1) is taken out (the circuit is made open) and magnetic field lines are drawn over the horizontal plane ABCD, the lines are
In Figure 13.1 a long straight conductor passes vertically through a horizontal card ABCD at the point O, in series with a battery, a variable resistance, an ammeter and a key. A compass is placed on the card.
Solution. The decisive words are “the key is taken out” — the circuit is open, so no current flows.
With no current there is no magnetic field due to the conductor at all. The concentric circles of options (a) and (d) are what you would draw with the key in; with it out they vanish entirely.
But the card is not in a field-free region. The Earth’s own magnetic field is still present, and over a region as small as a card it is essentially uniform, pointing from geographic south to north. A uniform field is drawn as straight, parallel, equidistant lines — which is option (c), and which is exactly what Question 1(c) has just established that parallel lines mean.
So the compass on the card simply settles along the north–south direction, as it would anywhere else.
This pairing of Questions 1 and 2 is neat: the first tells you what parallel lines represent, and the second gives a physical situation that produces them.
(c) straight lines parallel to each other — with the key out no current flows, so only the Earth’s uniform field remains
Question 3
A circular loop placed in a plane perpendicular to the plane of paper carries a current when the key is ON. The current as seen from points A and B (in the plane of paper and on the axis of the coil) is anti clockwise and clockwise respectively. The magnetic field lines point from B to A. The N-pole of the resultant magnet is on the face close to
Solution. A current-carrying circular loop behaves like a short bar magnet, with one face acting as a north pole and the other as a south pole.
The rule that identifies them is that field lines emerge from the north pole and enter the south pole.
The question states that the field lines point from B to A — that is, they travel through the loop towards A and come out of the face near A. Since lines emerge from a north pole, the N-pole is the face close to A. Option (a).
The same conclusion follows from the current directions given. Seen from A the current is anticlockwise, and an anticlockwise current seen from a face means that face is the north pole; seen from B it is clockwise, marking B’s face as the south pole. The two pieces of information agree, which is a useful internal check.
Options (c) and (d) suggest the polarity depends on the size of the current, which it does not. Increasing the current makes the field stronger, but the poles stay where they are. Only reversing the current would swap them — which is the point Question 4(d) makes about solenoids.
(a) A — the field lines emerge from the face near A, and lines emerge from a north pole
Question 4
For a current in a long straight solenoid N- and S-poles are created at the two ends. Among the following statements, the incorrect statement is
Solution. Three of these are true statements about solenoids; one contradicts the central fact of the topic.
(a) is true. Inside a long solenoid the field lines are straight, parallel and equally spaced, which by Question 1 means the field is uniform — the same in magnitude and direction at every interior point.
(b) is true, and it is how an electromagnet is made. A soft iron rod placed inside a current-carrying solenoid becomes strongly magnetised, and loses its magnetism when the current stops. Question 14 in the short-answer band asks how to make the magnetism permanent instead.
(d) is true. Reversing the current reverses the field direction throughout, so the two ends swap their polarity. This is precisely why an electromagnet’s poles can be switched at will while a permanent magnet’s cannot.
(c) is false, and is the answer. The field pattern of a solenoid is almost identical to that of a bar magnet — the same shape of lines emerging from one end, looping round outside and entering the other. That similarity is the whole reason a solenoid can be described as having N- and S-poles at all, and it is the strongest evidence that the magnetism of a bar magnet also arises from circulating currents.
(c) — the solenoid’s field pattern is the same as a bar magnet’s, not different from it
Question 5
A uniform magnetic field exists in the plane of paper pointing from left to right as shown in Figure 13.3. In the field an electron and a proton move as shown. The electron and the proton experience
Solution. Two things are reversed between the two particles — the direction of motion and the sign of the charge — and the key to the question is that two reversals cancel.
Take axes in the plane of the page: $\hat{x}$ to the right, $\hat{y}$ up the page, and $\hat{z}$ out of the page. The field is $\vec{B} = B\hat{x}$.
The proton moves up, $\vec{v} = v\hat{y}$, with charge $+e$:
$$\vec{F} = q\,\vec{v} \times \vec{B} = (+e)(v\hat{y}) \times (B\hat{x}) = evB\,(\hat{y} \times \hat{x}) = -evB\,\hat{z}$$
The minus sign on $\hat{z}$ means the force points into the page.
The electron moves down, $\vec{v} = -v\hat{y}$, with charge $-e$:
$$\vec{F} = (-e)(-v\hat{y}) \times (B\hat{x}) = (+evB)(\hat{y} \times \hat{x}) = -evB\,\hat{z}$$
Again into the page.
So both forces point into the plane of the paper — option (a).
The result can be seen without algebra once the cancellation is noticed: reversing the velocity reverses the force, and reversing the charge reverses it again, so two reversals leave the direction unchanged. Option (c), which assumes only one reversal, is the trap.
Option (d) is impossible for a different reason worth stating: the magnetic force is always perpendicular to both $\vec{v}$ and $\vec{B}$, so it can never point along or against the field.
(a) forces both pointing into the plane of paper — the reversed velocity and the reversed charge cancel each other
Question 6
Commercial electric motors do not use
Solution. The question asks what commercial motors do not use, so three options describe real features and one describes what only a toy motor has.
(a) Commercial motors do use an electromagnet rather than a permanent magnet, because an electromagnet can be made far stronger and its strength can be controlled.
(b) They do use a large number of turns, because the force on the coil — and so the torque — is proportional to the number of turns.
(d) They do wind the coil on a soft iron core. The core concentrates the magnetic field lines through the coil and greatly increases the field, and the coil and core together are called the armature.
(c) is therefore the answer: commercial motors do not use a permanent magnet. The simple demonstration motor of Question 29 uses one, which is exactly the difference between the two that question asks about.
The three features of (a), (b) and (d) all point the same way — every one of them is a method of getting more torque out of the same current, which is what a motor doing real work requires.
(c) a permanent magnet to rotate the armature — commercial motors use an electromagnet instead, along with many turns and a soft iron core
Question 7
In the arrangement shown in Figure 13.4 there are two coils wound on a non-conducting cylindrical rod. Initially the key is not inserted. Then the key is inserted and later removed. Then
In Figure 13.4 the primary coil is connected to a battery and a key, and the secondary coil, wound on the same non-conducting rod but with no electrical connection to the first, is connected to a galvanometer.
Solution. This is the classic demonstration of electromagnetic induction, and the governing fact is that a current is induced only while the magnetic field through the secondary coil is changing.
When the key is inserted, the current in the primary rises from zero to its steady value. The field through the secondary therefore increases, and while it is increasing a current is induced — the galvanometer gives a momentary deflection. Once the current is steady the field stops changing and the deflection dies out.
When the key is removed, the primary current falls back to zero, so the field through the secondary decreases. That is again a change, so again a momentary deflection is seen — and because the field is now changing in the opposite sense, the induced current flows the other way and the needle deflects in the opposite direction.
So there are two momentary deflections in opposite directions, which is option (d).
Option (a) misses induction entirely; option (b) forgets that switching off is as much a change as switching on; option (c) gets the deflections but not their reversal — and the reversal is the whole evidence that it is the direction of the change that matters, not the presence of a current.
Nothing happens while the current is merely flowing steadily, which is the central and often surprising point: a steady field induces nothing.
(d) momentary deflections in opposite directions — one as the field grows when the key is inserted, one as it collapses when the key is removed
Question 8
Choose the incorrect statement
Solution. Statements (a), (b) and (c) are correct as written — (a) and (b) are the rules in the table at the top of this page, and (c) is the standard definition of DC and AC.
(d) is wrong, and the error is a factor of two.
The frequency of AC in India is 50 Hz, which means the current completes 50 full cycles per second. But in each cycle the current reverses direction twice — once going positive to negative and once coming back. So the number of reversals per second is
$$2 \times 50 = 100$$
and the time between successive reversals is
$$\frac{1}{100}\ \mathrm{second}, \ \text{not } \frac{1}{50}$$
The $\frac{1}{50}$ second is the period of one complete cycle, not the interval between direction changes.
This is a standard trap, and the way to avoid it is to picture one cycle: the current rises, falls through zero (first reversal), goes negative, and rises back through zero (second reversal). Two crossings per cycle, hence 100 per second. Question 23 in the short-answer band asks for the same number directly.
(d) — AC in India changes direction every $\frac{1}{100}$ second, not $\frac{1}{50}$, because each 50 Hz cycle contains two reversals
Question 9
A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in Figure 13.5. The direction of magnetic field at a point will be North to South
Solution. The wire lies in the plane of the paper running east–west, so “above” and “below” in options (a) and (b) mean out of and into the page, while (c) and (d) mean north and south within the page.
By the right-hand thumb rule, point the right thumb along the current — towards the west — and the fingers curl in the direction of the field. The field lines are circles round the wire.
Working it with components makes it certain. Let $\hat{x}$ be east, $\hat{y}$ north and $\hat{z}$ out of the page. The current is along $-\hat{x}$, and the field at a point in the direction $\hat{r}$ from the wire goes as $\hat{I} \times \hat{r}$.
Directly below the wire, meaning behind the page, $\hat{r} = -\hat{z}$:
$$(-\hat{x}) \times (-\hat{z}) = \hat{x} \times \hat{z} = -\hat{y}$$
which points south. So there the field runs from north to south — option (b).
Directly above the wire, $\hat{r} = +\hat{z}$, gives $(-\hat{x}) \times \hat{z} = +\hat{y}$, pointing north — the opposite sense, so (a) is wrong.
At points in the plane of the paper, north or south of the wire, the circular field lines are perpendicular to the page, so the field points out of or into the page — not north–south at all. That disposes of (c) and (d).
(b) directly below the wire
Question 10
The strength of magnetic field inside a long current carrying straight solenoid is
Solution. Inside a long solenoid the field lines run straight, parallel and equally spaced along the axis. By the spacing convention of Question 1, equally spaced lines mean a field of constant strength — the field is uniform, the same at every interior point. Option (c).
The physical reason is that a point well inside a long solenoid has a great many turns on either side of it, all contributing, and moving along the axis merely exchanges turns on one side for turns on the other. The total stays the same.
The word long is doing real work. The uniformity holds well away from the ends; near the ends the field does weaken and the lines spread out, which is exactly what Question 20 asks about. Option (a) states the reverse of that, and (b) and (d) describe variations that do not occur.
This uniform interior field is what makes a solenoid so useful: it is the standard way of producing a known, even magnetic field, and it is why a soft iron rod placed inside is magnetised evenly along its length to make an electromagnet.
(c) same at all points — the field inside a long solenoid is uniform, shown by straight, parallel, equally spaced field lines
Question 11
To convert an AC generator into DC generator
Solution. In both kinds of generator the coil rotates in a magnetic field and the induced current in the coil reverses every half turn — that much is unavoidable, because after half a revolution each side of the coil has changed places with the other.
The difference is entirely in how the coil is connected to the external circuit.
An AC generator uses slip rings: each end of the coil is permanently attached to its own continuous ring, so the reversal inside the coil is passed straight out. The external current alternates.
A DC generator uses a split-ring commutator: a single ring cut into two halves, each connected to one end of the coil. The brushes are fixed, and at the very moment the current in the coil reverses, the two halves of the ring exchange brushes. The reversal inside the coil is therefore cancelled by a reversal of the connections, and the current in the external circuit always flows the same way.
So option (a). Option (b) describes the AC generator itself, so it is the opposite of what is asked. Options (c) and (d) would change the size of the induced current, not its direction — a stronger field or a different coil shape gives more current, still alternating.
The same split-ring commutator appears in the motor of Question 29, doing the mirror-image job: there it reverses the current in the coil so that the rotation continues in one sense.
(a) split-ring type commutator must be used — it reverses the connections at the instant the coil’s current reverses, so the external current stays one-way
Question 12
The most important safety method used for protecting home appliances from short circuiting or overloading is
Solution. Read what is being protected and against what: appliances, against short circuiting or overloading — that is, against excessive current.
A fuse is a thin wire of low melting point in series with the circuit. When the current exceeds its rated value the heat produced ($I^2Rt$) melts it, the circuit breaks, and the current stops before the appliance or the wiring is damaged. That is precisely the stated job, so option (b).
The other three all do useful things, but not this one.
Earthing connects the metal body of an appliance to the ground through a low-resistance wire. Its purpose is to protect the user from electric shock if a live wire touches the casing — it safeguards people, not appliances, and it does not act on overload.
A stabiliser protects against voltage fluctuations, holding the supply near its nominal value. A short circuit is a current problem, not a voltage one.
An electric meter measures the energy consumed for billing. It is a measuring instrument with no protective function at all.
The distinction between the fuse and earthing is the one worth carrying, and it is a common examination point: the fuse protects the appliance; earthing protects the person.
(b) use of fuse — it melts and breaks the circuit when the current exceeds its rating
Common mistakes
- Reading parallel field lines as zero field, in Questions 1 and 2. Parallel, equidistant lines mean a uniform field. Zero field is shown by no lines at all.
- Applying only one reversal in Question 5. The electron has both the opposite velocity and the opposite charge, so the two reversals cancel and both particles are pushed the same way.
- Using $\frac{1}{50}$ second in Question 8. That is the period of one cycle; the current reverses twice per cycle, so the interval between reversals is $\frac{1}{100}$ second.
- Thinking a steady current induces something, in Question 7. Induction requires a changing field. Deflections occur only while the primary current is growing or collapsing, and they are in opposite directions.
- Mixing up the three hand rules, in Question 8. Right-hand thumb gives the field round a current; Fleming’s left hand gives the force in a motor; Fleming’s right hand gives the induced current in a generator.
- Saying the coil’s polarity depends on the current’s size, in Question 3. More current gives a stronger field, not a different pole. Only reversing the current swaps N and S.
- Confusing earthing with the fuse, in Question 12. The fuse protects the appliance from excess current; earthing protects the user from shock.
- Choosing slip rings for DC, in Question 11. Slip rings give AC; the split-ring commutator is what makes the output one-way.
Practise next
- Exemplar Short and Long Answer Questions — where the field patterns, the motor and the generator have to be drawn and explained rather than recognised.
- NCERT textbook Chapter 12 — Magnetic Effects of Electric Current for the end-of-chapter exercise on field patterns, the motor rule and induction.
- Exemplar Electricity — the previous unit, whose currents and circuits this one builds directly on, including the fuse of Question 12.
- Textbook activities for the compass and iron-filing experiments behind Questions 2, 9 and 10.