NCERT Class 12 Physics — Electromagnetic Induction, Chapter 6 Exercises. All 8 questions solved.

The exercises at the end of Chapter 6 fall into two groups. Questions 6.1 and 6.2 ask only for the direction of an induced current, and are answered with Lenz’s law and no arithmetic at all. Questions 6.3 to 6.8 ask for the size of an induced emf or an inductance, and every one of them is Faraday’s law applied to a flux that changes for a different reason — because the field changes, because the area changes, or because a conductor sweeps across field lines. The results used are:

• Faraday’s law — the induced emf is $\varepsilon = -\dfrac{\text{d}\Phi_B}{\text{d}t}$, where $\Phi_B = \vec B\cdot\vec A$ is the magnetic flux through the circuit (times the number of turns, for a coil).
• Lenz’s law — the induced current flows in the direction that opposes the change of flux producing it. That is the minus sign in Faraday’s law.
• Motional emf and inductance — a rod of length $l$ moving at speed $v$ at right angles to a field $B$ has an emf $\varepsilon = Blv$ across it; a changing current induces $\varepsilon = -L\,\dfrac{\text{d}I}{\text{d}t}$ in its own circuit and $\varepsilon_2 = -M\,\dfrac{\text{d}I_1}{\text{d}t}$ in a neighbouring one.

Key insight. For every direction question, ask first whether the flux through the loop is growing or shrinking, and only then look at the loop. An induced current always tries to keep the flux as it was: against a growing flux it sets up an opposing field, and when the flux is dying away it sets up a field in the same direction to prop it up. That one rule settles all eight situations in 6.1 and 6.2, including the two that trip students up — the current that flows the same way as the primary current when a key is released (6.1(e)), and the loop in which no current flows at all (6.1(f)).

Question 6.1

Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).

How each part is worked. Decide whether the flux through the loop or coil is increasing or decreasing, then apply Lenz’s law to find the direction of the field the induced current must produce. To turn that into a direction round the circuit, use the rule for a current loop: seen from one face, a current circulating anticlockwise makes that face a north pole, and a current circulating clockwise makes it a south pole. For the coils in (a) and (b), which lead the current leaves by also depends on the sense in which the coil is wound; the paths given below read the winding as the book’s answers do, with each lead passing over (or under) the coil into the far side of its first turn.

Question 6.1 (a)

Solution. The magnet moves towards the coil with its south pole leading, so the flux through the coil increases. To oppose the approach, the face of the coil nearest the magnet must become a south pole and push the magnet back. Seen from the magnet, then, the induced current circulates clockwise. With the coil wound as drawn, that current flows through the coil from $p$ to $q$ and back round the external circuit from $q$ through $r$ to $p$.

Question 6.1 (b)

Solution. The magnet is moving towards the left-hand coil and away from the right-hand coil, so the two coils see opposite changes.

Left coil, $pqr$. The south pole is approaching, so the flux increases. The near face of the coil becomes a south pole to repel it, and seen from the magnet the current circulates clockwise. With the winding drawn, it flows through the coil from $q$ to $p$, and round the circuit from $p$ through $r$ to $q$.

Right coil, $xyz$. The north pole is moving away, so the flux decreases. To oppose the retreat, the near face of the coil becomes a south pole, attracting the north pole back. Seen from the magnet the current again circulates clockwise; through the coil it flows from $x$ to $y$, and round the circuit from $y$ through $z$ to $x$.

Question 6.1 (c)

Solution. Closing the tapping key starts a current in the left-hand loop, in the direction of the arrows: up its left side and down its right side as drawn. The current grows from zero, so the flux it sends along the common axis through loop $xyz$ grows too. The induced current in $xyz$ opposes that growth, so it circulates in the opposite sense to the current in the first loop — down its left side and up its right side, from $y$ to $z$ to $x$.

Question 6.1 (d)

Solution. The arrow at the rheostat shows its sliding contact moving towards the end of the resistance wire that is connected into the circuit. Less of the wire is then in use, so the resistance falls and the current in the right-hand loop grows. That current flows up the loop’s right side and down its left side, as the arrows show. The flux through loop $xyz$ grows, so its induced current circulates the opposite way — up its left side and down its right side, from $z$ to $y$ to $x$.

Question 6.1 (e)

Solution. While the key is held down, a steady current flows round the first solenoid’s circuit. The longer plate of the cell, its positive terminal, faces the key, so the current flows through the key, into the right-hand end of the solenoid, and through it from right to left. Releasing the key stops this current, so the flux through the second solenoid falls. The induced current opposes the fall by trying to keep the flux as it was, which means it flows in the same sense as the current that has just stopped. The two solenoids are wound alike, so in the second one the current also flows from right to left, from $y$ to $x$, and returns through the external circuit from $x$ through $r$ to $y$.

Question 6.1 (f)

Solution. The field lines of a straight current are circles centred on the wire, lying in planes at right angles to it. Here the wire passes through the centre of the loop at right angles to its plane, so the field lines lie in the plane of the loop and not one of them passes through it. The flux through the loop is zero whatever the current is, so it stays zero while the current decreases, and nothing is induced.

Question 6.2

Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.

Question 6.2 (a)

Solution. The crosses show a field directed into the page. As the irregular loop pulls out into a circle (the arrows at $a$ and $c$ show the wire moving outwards), the area it encloses grows: of all the shapes a given length of wire can make, the circle encloses the largest area. So the flux into the page increases. The induced current opposes the increase by producing a field out of the page inside the loop, which means it circulates anticlockwise as seen in the figure: from $a$ to $d$ to $c$ to $b$.

Note on the book’s answer. The key prints “Along adcd”. A path round the loop has to pass all four labelled points, and “adcd” comes back to $d$ without reaching $b$. It is a misprint for $adcb$, the anticlockwise sense found above; the key’s own explanation, that the flux increases and the induced current opposes it, gives the same.

Question 6.2 (b)

Solution. Here the dots show a field directed out of the page. Squeezing the circle into a narrow straight wire — $a$ and $c$ are pushed in to $a’$ and $c’$ while $b$ and $d$ spread out to $b’$ and $d’$ — shrinks the enclosed area almost to nothing, so the flux out of the page decreases. The induced current opposes the decrease by producing a field out of the page inside the loop, to make up for the flux being lost. That is an anticlockwise current as seen in the figure: $a’$ to $d’$ to $c’$ to $b’$.

Question 6.3

A long solenoid with $15$ turns per cm has a small loop of area $2.0\ \text{cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0\ \text{A}$ to $4.0\ \text{A}$ in $0.1\ \text{s}$, what is the induced emf in the loop while the current is changing?

Solution. Inside a long solenoid the field is uniform and along the axis, $B = \mu_0nI$, with $n = 15$ turns per cm $= 1500$ turns per metre. The loop is normal to the axis, so the field passes straight through it and the flux is $\Phi = BA = \mu_0nIA$. Only the current changes, so

$$\varepsilon = \frac{\text{d}\Phi}{\text{d}t} = \mu_0nA\,\frac{\text{d}I}{\text{d}t} = (4\pi\times10^{-7})(1500)(2.0\times10^{-4})\left(\frac{4.0 – 2.0}{0.1}\right) = 7.5\times10^{-6}\ \text{V}$$

The area in the flux is the loop’s own $2.0\ \text{cm}^2 = 2.0\times10^{-4}\ \text{m}^2$, not the cross-section of the solenoid, because the flux that matters is the flux through the loop.

Question 6.4

A rectangular wire loop of sides $8\ \text{cm}$ and $2\ \text{cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3\ \text{T}$ directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1\ \text{cm s}^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Solution. While the loop is crossing the edge of the field region, the part of it still inside the field shrinks, and so does the flux through it. If $l$ is the side parallel to the edge (the side at right angles to the velocity), the area inside the field falls by $lv$ every second, so

$$\varepsilon = Blv$$

This is the motional emf of the side still in the field, which is the only side cutting field lines. It lasts until the whole loop is out, that is, while the loop moves a distance equal to its other side. Throughout, $v = 1\ \text{cm s}^{-1} = 0.01\ \text{m s}^{-1}$.

Question 6.4 (a)

Solution. The velocity is normal to the longer side, so the $8\ \text{cm}$ side lies along the edge of the field and $l = 0.08\ \text{m}$:

$$\varepsilon = (0.3)(0.08)(0.01) = 2.4\times10^{-4}\ \text{V}$$

The loop leaves the field after moving its shorter side, $2\ \text{cm}$, which at $1\ \text{cm s}^{-1}$ takes $2\ \text{s}$.

Question 6.4 (b)

Solution. Now the $2\ \text{cm}$ side lies along the edge, so $l = 0.02\ \text{m}$:

$$\varepsilon = (0.3)(0.02)(0.01) = 0.6\times10^{-4}\ \text{V}$$

The loop has to move its longer side, $8\ \text{cm}$, to get clear, which takes $8\ \text{s}$. The emf is a quarter as large but lasts four times as long, so the total change of flux, $B\times$ area, is the same either way.

Question 6.5

A $1.0\ \text{m}$ long metallic rod is rotated with an angular frequency of $400\ \text{rad s}^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of $0.5\ \text{T}$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Solution. Different parts of the rod move at different speeds: a small piece of length $\text{d}r$ at distance $r$ from the axis moves at $v = \omega r$, at right angles to the field, and so has a motional emf $\text{d}\varepsilon = Bv\,\text{d}r = B\omega r\,\text{d}r$ across it. These pieces are in series along the rod, so their emfs add:

$$\varepsilon = \int_0^{l} B\omega r\,\text{d}r = \frac12B\omega l^2 = \frac12(0.5)(400)(1.0)^2 = 100\ \text{V}$$

The same result comes from the flux: the rod sweeps out area at the rate $\tfrac12\omega l^2$, and $B$ times that rate is the emf.

Question 6.6

A horizontal straight wire $10\ \text{m}$ long extending from east to west is falling with a speed of $5.0\ \text{m s}^{-1}$, at right angles to the horizontal component of the earth’s magnetic field, $0.30\times10^{-4}\ \text{Wb m}^{-2}$. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?

Question 6.6 (a)

Solution. The wire, its velocity and the horizontal field are mutually perpendicular — east-west, downwards and south-north — so the motional emf is simply

$$\varepsilon = Blv = (0.30\times10^{-4})(10)(5.0) = 1.5\times10^{-3}\ \text{V}$$

The vertical component of the earth’s field plays no part, because the wire moves along it.

Question 6.6 (b)

Solution. The direction of the emf is the direction of the magnetic force $q\,\vec v\times\vec B$ on a positive charge in the wire. The horizontal field points north and the velocity points down. Taking east, north and up as the $x$, $y$ and $z$ directions, $\vec v\times\vec B$ is along $(-\hat k)\times\hat{\jmath} = \hat{\imath}$, which is east. Fleming’s right-hand rule gives the same: first finger north, thumb down, second finger east.

Question 6.6 (c)

Solution. The emf pushes positive charge towards the eastern end of the wire, so the eastern end becomes positive relative to the western end.

Question 6.7

Current in a circuit falls from $5.0\ \text{A}$ to $0.0\ \text{A}$ in $0.1\ \text{s}$. If an average emf of $200\ \text{V}$ is induced, give an estimate of the self-inductance of the circuit.

Solution. The self-induced emf is proportional to the rate of change of the current, $|\varepsilon| = L\,|\text{d}I/\text{d}t|$. The current falls by $5.0\ \text{A}$ in $0.1\ \text{s}$, a rate of $50\ \text{A s}^{-1}$, so

$$L = \frac{|\varepsilon|}{|\text{d}I/\text{d}t|} = \frac{200}{50} = 4\ \text{H}$$

It is an estimate because only the average rate of change of the current is known.

Question 6.8

A pair of adjacent coils has a mutual inductance of $1.5\ \text{H}$. If the current in one coil changes from $0$ to $20\ \text{A}$ in $0.5\ \text{s}$, what is the change of flux linkage with the other coil?

Solution. Mutual inductance is defined by the flux linkage one coil’s current produces in the other, $N\Phi = MI$. The change of flux linkage therefore depends only on the change of current, not on how quickly it happens:

$$\Delta(N\Phi) = M\,\Delta I = (1.5)(20) = 30\ \text{Wb}$$

The $0.5\ \text{s}$ would be needed only for the emf induced in the second coil, $M\,\Delta I/\Delta t = 60\ \text{V}$.

Common mistakes

• Question 6.1(a) and (b): getting the pole the wrong way round. An approaching pole always finds a like pole facing it (the coil repels it), and a retreating pole always finds an unlike pole (the coil attracts it back). Students who make the near face a north pole in (a) because the flux is increasing have forgotten that it is a south pole that is approaching.
• Question 6.1(e): assuming the induced current always opposes the primary current. It opposes the change in flux. When the key is released the flux is dying away, so the induced current flows the same way as the primary current did, to prop the flux up.
• Question 6.1(f): predicting a current because the current is changing. A changing current induces nothing unless flux through the loop changes. Here the field lines lie in the loop’s plane, so the flux is zero throughout.
• Question 6.2(a): thinking the area cannot change because the wire’s length does not. A fixed length of wire encloses more area as a circle than in any other shape, so pulling it into a circle increases the flux.
• Question 6.3: using the wrong $n$ or the wrong area. $15$ turns per cm is $1500$ turns per metre, not $15$. And the area is that of the small loop, because it is the flux through the loop that induces its emf.
• Question 6.4: swapping the durations. The emf is set by the side parallel to the edge of the field, but the time is set by the other side — the distance the loop must travel to get out. That is why the larger emf in (a) lasts the shorter time.
• Question 6.5: using $Blv$ with the speed of the tip. The tip moves at $\omega l = 400\ \text{m s}^{-1}$, but the rest of the rod moves more slowly, down to zero at the axis. Using the tip speed gives $200\ \text{V}$, twice the answer; the average speed along the rod is $\omega l/2$.
• Question 6.8: dividing by the time. $M\,\Delta I/\Delta t = 60\ \text{V}$ is the induced emf. The change of flux linkage asked for is $M\,\Delta I = 30\ \text{Wb}$.

Practise next

• Chapter 7, Alternating Current — the self-inductance of 6.7 reappears as the reactance of an inductor in an ac circuit.
• Chapter 4, Moving Charges and Magnetism — the fields of a straight wire and a solenoid used in 6.1(f) and 6.3.