Chapter 13 · Magnetic Effects of Electric Current

NCERT Exemplar Class 10 Science Chapter 13 Long Answer Questions

All 8 questions of NCERT Exemplar Class 10 Science Chapter 13 Long Answer Questions (Magnetic Effects of Electric Current), in the book's order, each with its answer and a step-by-step solution.

1. Q25
   Why does a ​magnetic compass needle pointing North and South in the absence of a nearby magnet get deflected when a bar magnet or ​a current ​carrying loop ​is brought ​near it. Describe ​some salient features of magnetic ​lines of field ​concept.
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   The needle sets along the resultant of the Earth's field and the stronger field of the magnet or loop; field lines are closed curves.

   A compass needle is itself a tiny bar magnet, pivoted so that it can turn freely. When no magnet is near, the only field acting on it is the Earth's weak, nearly horizontal field, so the couple on the needle turns it until it lies along that field, pointing north and south. Bringing a bar magnet near, or a loop carrying a current, adds a second and much stronger magnetic field at the needle. The needle now sets itself along the **resultant** of the two fields, and so is deflected away from the north-south line; with the magnet's field equal to the Earth's and at right angles to it, the deflection is $45^{\circ }$. Salient features of magnetic field lines. First, the tangent to a line at any point gives the direction of the magnetic field there, which is the direction in which the north pole of a compass needle placed at that point sets itself. Second, outside a magnet the lines run from the north pole to the south pole and inside the magnet from south to north, so every field line is a closed continuous curve with no beginning and no end. Third, no two field lines ever cross, because the field cannot have two directions at one point. Fourth, the relative strength of the field is shown by the closeness of the lines: crowded lines, as near the poles, mean a strong field, widely spaced lines a weak one, and parallel equidistant lines a uniform field.

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2. Q26
   With the ​help of a labelled circuit diagram ​illustrate ​the pattern of field lines of the magnetic field around ​a current carrying straight ​long conducting wire. How is the ​right hand thumb ​rule useful to find direction of ​magnetic field associated with a current carrying conductor?
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   The field lines are concentric circles around the wire, and the right-hand thumb rule, with the thumb along the current, gives their direction.

   The pattern. A long straight thick wire XY is passed vertically through a hole at the centre of a horizontal cardboard sheet and joined in series with a battery, a plug key, a rheostat and an ammeter. Iron filings are sprinkled on the cardboard and the key is plugged in. On tapping the board the filings settle into **concentric circles** centred on the wire, and every small compass needle placed on the board sets itself along the tangent to one of these circles. So the magnetic field lines around a long straight current-carrying conductor are concentric circles centred on the conductor and lying in a plane at right angles to it. Moving outward the circles are farther and farther apart, showing that the field weakens with distance, and increasing the current makes the pattern stronger and reach farther, so that $B\propto I/r$. The rule. The right-hand thumb rule fixes the **direction** of that field. Imagine holding the current-carrying straight conductor in your right hand with the thumb pointing along the direction of the conventional current; the fingers wrapped round the conductor then point along the direction of the magnetic field lines. Hence a current coming out of the plane of the paper is encircled by anticlockwise field lines, and a current going into the paper by clockwise ones; reversing the current reverses the sense of every circle without moving it.

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3. Q27
   Explain with ​the help of a labelled diagram the ​distribution of magnetic field due ​to a current through a circular ​loop. Why is it that if ​a current carrying coil has $n$ turns the field produced at any point is $n$ times as large as ​that ​produced by a single turn?
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   Field lines near the wire are circles, straight at the centre; the $n$ equal fields of the series turns add up to $n$ times.

   Distribution of the field. Every short piece of the circular wire acts as a straight current-carrying conductor, so by the right-hand thumb rule it is encircled by magnetic field lines which, close to that piece, are almost circles centred on it. As we move from the wire towards the centre of the loop these circles become larger and larger, until at the centre of the loop their arcs appear as a straight line at right angles to the plane of the loop. Because the current runs round the loop in one consistent sense, every piece of the wire drives the field through the loop in the **same** direction, so the contributions add: the field is strongest at the centre and weakens as we move away. The loop therefore behaves as a small magnetic dipole, with the face from which the lines emerge acting as a north pole and the face they enter as a south pole; seen from a face, an anticlockwise current makes that face a north pole. For a single loop of radius $r$ carrying current $I$ the field at the centre is $B=\frac{{\mu }_{0}I}{2r}$. Why n turns give n times the field. In a coil of $n$ turns the turns are joined end to end, that is, in series, so the **same** current $I$ flows through each of them, and each turn has nearly the same radius and is wound the same way round. Each turn therefore produces at the centre a field equal in size and identical in direction to that of any other turn. Magnetic fields obey the principle of superposition, so the resultant is the vector sum of the $n$ contributions; since they are all parallel, the vector sum is simply the arithmetic sum. Hence $B=n{B}_1=\frac{{\mu }_{0}nI}{2r}$, that is, $n$ times the field of a single turn.

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4. Q28
   Describe the activity that shows that a ​current-carrying conductor experiences a force ​perpendicular to its length and the external magnetic ​field. How does Fleming’s left-hand rule ​help us to find ​the direction ​of the force acting on the current carrying ​conductor?
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   A current-carrying rod between magnet poles is pushed sideways, at right angles to the rod and the field; Fleming's left-hand rule gives the direction.

   The activity. Take a small aluminium rod AB about $5$ cm long and suspend it horizontally from a stand by two flexible connecting wires. Place a strong horseshoe magnet so that the rod lies between its poles, with the north pole vertically above the rod and the south pole below it, so that the magnetic field at the rod is directed vertically downward. Connect the rod in series with a battery, a plug key and a rheostat. Now pass a current through the rod from end B to end A: the rod is displaced sideways, to the left, at right angles both to its own length and to the field. Reverse the direction of the current and the rod is displaced to the right. Restore the current and instead interchange the poles of the magnet, so that the field is reversed: again the displacement reverses. The displacement is largest when the rod is at right angles to the field and falls to zero when the rod is turned to lie along the field. This shows that a current-carrying conductor placed in an external magnetic field experiences a force perpendicular to its own length and to the field, whose direction is reversed by reversing either the current or the field, so that $F=BIL\sin \theta$. Fleming's left-hand rule. Stretch the thumb, the first finger and the centre finger of the **left** hand so that all three are mutually at right angles. If the first finger points in the direction of the magnetic field and the centre finger in the direction of the current, then the thumb points in the direction of the force, that is, of the motion of the conductor. Applied to the arrangement above, the first finger points down, the centre finger points from B to A into the page, and the thumb points to the left, which is exactly the direction in which the rod is seen to move.

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5. Q29
   Draw ​a labelled circuit diagram of a simple electric motor ​and ​explain its ​working. In what way ​these simple electric motors are diffferent from ​commercial motors?
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   The forces on the coil arms form a couple that turns the coil; the split ring keeps it turning, and commercial motors use electromagnets.

   Construction. A simple electric motor has a rectangular coil ABCD of insulated copper wire placed between the poles of a strong permanent magnet, with the arms AB and CD at right angles to the field. The ends of the coil are joined to the two halves P and Q of a split ring fixed to the axle; two stationary brushes X and Y press on P and Q and connect the coil, through a key, to a battery. Working. Current enters at brush X, passes through ring half P and round the coil along $A\to B\to C\to D$, and leaves through Q and brush Y. The current in arm AB is therefore opposite in direction to that in arm CD. Applying Fleming's left-hand rule, the force on AB is downward and the force on CD is upward; the arms BC and DA lie along the field and feel no force. These two equal and opposite forces form a couple, and the coil rotates anticlockwise. After half a rotation, ring half Q comes into contact with brush X and P with brush Y, so the current through the coil is reversed and now flows along $D\to C\to B\to A$. The force on each arm is reversed with it, so the arm now on the left is again pushed down and the one on the right pushed up, and the coil continues to turn in the **same** direction. Because the split ring reverses the current after every half rotation, the rotation is continuous; a device that does this is called a commutator. Differences from commercial motors. In a commercial motor an electromagnet is used in place of the permanent magnet, so that a much stronger field is available; the conducting coil has a very large number of turns; and the coil is wound on a soft iron core, the core and the coils together being called the armature, which greatly increases the magnetic field and hence the power of the motor.

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6. Q30
   Explain the ​phenomenon ​of electromagnetic induction. Describe an experiment ​to show that a ​current is set up in a closed ​loop when an external magnetic ​field ​passing through the loop ​increases or decreases.
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   A changing magnetic field through a closed loop induces a current, shown by a galvanometer when the key is plugged in or pulled out.

   Electromagnetic induction. Whenever the magnetic field passing through a closed conducting loop **changes**, an electromotive force is induced across the loop, and since the loop is closed an induced current flows in it. This is electromagnetic induction. The current flows only while the change is going on; a steady field, however strong, induces nothing. The field through the loop may be changed by moving a magnet towards or away from the loop, by moving the loop, by rotating it in a field, or simply by changing the current in a neighbouring coil. The direction of the induced current in a conductor moved across a field is given by Fleming's right-hand rule: with the first finger of the right hand along the field and the thumb along the motion, the centre finger gives the induced current. The experiment. Take two coils of copper wire with a large number of turns, say $50$ turns and $100$ turns, and insert them one over the other on a non-conducting cylindrical roll. Join coil-1 in series with a battery and a plug key, and join coil-2 to a galvanometer, so that coil-2 forms a closed loop. Plug in the key: the galvanometer needle gives a momentary deflection and returns to zero, because while the current in coil-1 is growing from zero the magnetic field passing through coil-2 is **increasing**. Keep the key plugged in so that a steady current flows: the needle stays at zero however long one waits, because the field through coil-2 is now steady. Now pull out the key: the needle gives another momentary deflection, this time in the **opposite** direction, because the current in coil-1 and the field through coil-2 are **decreasing** to zero. The same result is obtained more simply by pushing a bar magnet into a coil joined to a galvanometer, holding it still, and withdrawing it. This shows that a current is set up in a closed loop whenever the external magnetic field passing through the loop increases or decreases, and that it reverses when the change reverses.

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7. Q31
   Describe the working of an AC generator ​with the help of ​a ​labelled circuit diagram. What changes ​must be ​made in the arrangement ​to ​convert it to a DC generator?
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   Slip rings give alternating current because the induced current reverses every half turn; one split ring in their place gives direct current.

   Working of an AC generator. A rectangular coil ABCD of insulated copper wire is mounted on an axle between the poles of a strong permanent magnet, so that the arms AB and CD lie at right angles to the field. The two ends of the coil are joined to two **separate** complete rings, the slip rings $R_1$ and $R_2$, which are fixed to the axle and turn with the coil; two stationary brushes $B_1$ and $B_2$ press on them and lead to the external circuit. When the axle is turned by an outside agency, the arm AB moves up across the field while CD moves down. By Fleming's right-hand rule the induced current runs from A to B in one arm and from C to D in the other, that is, the same way round the loop, so the two add and the current in the coil flows along $A\to B\to C\to D$ and out through $R_2$ and $B_2$. After half a rotation the two arms exchange places, so AB is now moving down and CD up; the induced current in each arm is reversed and the current in the coil flows along $D\to C\to B\to A$. Because each slip ring is a complete ring, every brush stays joined to the same end of the coil throughout, so this reversal is passed straight to the external circuit. The current therefore changes direction once every half rotation: it is an alternating current. In India the coil is turned $50$ times a second, so the current reverses $100$ times a second, that is once every $\frac{1}{100}$ s, and completes one cycle in $\frac{1}{50}$ s. Change needed for a DC generator. Replace the two slip rings by a **single split ring**, that is, by a commutator of the kind used in the electric motor, with one half joined to each end of the coil and the two brushes pressing on it. As the coil turns, each brush leaves one half and meets the other at exactly the instant at which the current in the coil reverses, so the reversal inside the coil is cancelled by the change of contact outside it. One brush then remains always in contact with the arm that is moving up and the other with the arm moving down, and the current in the external circuit flows in one direction only: it is a direct current. The coil, the magnet and the brushes are unchanged, so the split ring is the only change required.

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8. Q32
   Draw ​an appropriate schematic diagram showing ​common domestic circuits ​and ​discuss the ​importance ​of fuse. Why is it that a burnt out fuse should be replaced by another fuse of ​identical rating?
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   A fuse melts to break the circuit on excess current; a higher rating gives no protection and a lower one melts at normal current.

   The schematic. Electricity reaches a house through a main supply of three wires: the **live** wire, usually with red insulation, at a high potential; the **neutral** wire, usually black, at earth potential; and the **earth** wire, usually green, joined to a metal plate buried in the ground near the house. In India the potential difference between live and neutral is $220$ V. The supply passes through the electricity meter, then through a main fuse fitted in the live wire, then through the main switch, and so to the distribution box. From there two separate circuits are usually taken: one of $5$ A rating for bulbs, fans and other light appliances, and one of $15$ A rating for heavy appliances such as geysers and heaters. Each circuit carries its own fuse in the live wire. Every appliance is connected in **parallel** between the live and the neutral wire, so that each receives the full $220$ V and has its own switch, which is also placed in the live wire; the metal body of an appliance is joined to the earth wire. Importance of the fuse. A fuse is a short piece of wire of high resistance and low melting point, an alloy of tin and lead, joined in series in the live wire. In a short circuit the live and neutral wires touch and the resistance falls almost to zero, and in overloading too many appliances are run from one circuit; in both cases the current becomes far larger than normal. Since the heat produced goes as the square of the current, $H=I^{2}Rt$, this excessive current heats the thin fuse wire until it melts and breaks the circuit, before the wiring can overheat, the appliances be damaged or a fire start. The fuse is therefore the most important safety device in a domestic circuit. The earth wire is a separate safeguard: it carries a leakage current from the metal body of an appliance safely to the ground and so protects the user from a shock. Why the replacement must have the identical rating. The rating of a fuse is chosen just above the largest current the wiring and appliances of that circuit can safely carry. If a blown $5$ A fuse is replaced by a $15$ A one, a fault current of $12$ A will no longer melt it, although the wiring was chosen for $5$ A; the heat would then be produced at $(12/5{)}^2\approx 5.8$ times the intended rate, melting the insulation and possibly starting a fire, so the protection is lost at exactly the moment it is needed. If instead a fuse of lower rating is used, say $2$ A in the $5$ A lighting circuit, it melts at the perfectly normal working current of about $3$ A and the circuit keeps breaking although nothing is wrong. Only a fuse of the same rating both protects the circuit at the current at which it becomes unsafe and allows it to carry its normal working current.

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