Chapter 6: Electromagnetic Induction

In Experiment 6.2, two coils $$C_1$$ and $$C_2$$ are placed close together. When the current in one coil is changed, the changing magnetic flux through the other coil induces an emf, and the galvanometer deflects. The deflection is proportional to the induced emf, which equals the rate of change of flux linkage $$N\dfrac{d\Phi_B}{dt}$$.

To obtain a large deflection, this rate of change of flux linkage must be made as large as possible. One or more of the following steps can be taken:

• Place a rod of soft iron inside the coil $$C_2$$. The soft-iron core greatly increases the magnetic field and hence the flux linked with the coils.
• Connect the coil to a powerful battery, so that a large current flows and produces a strong magnetic field.
• Move the arrangement of the two coils rapidly towards the test coil $$C_1$$, so that the flux changes in a very short time, making $$\dfrac{d\Phi_B}{dt}$$ large.

The galvanometer is only a detector of the induced current. Its role can be played by any device that visibly responds to a current.

Replace the galvanometer by a small bulb, of the kind used in a small torch light. Now move the two coils rapidly relative to each other (or rapidly switch the current in the other coil on and off). The changing flux induces a current large enough to make the bulb glow. The glowing of the bulb demonstrates the presence of the induced current even though no galvanometer is used.

No. The loop is held stationary in a steady magnetic field. An emf (and hence a current) is induced only when the magnetic flux linked with the loop changes with time.

Here, neither the loop nor the magnets move, and the field of the permanent magnets is constant. So the flux through the loop does not change, and no current is induced.

Using a stronger magnet only makes the (constant) flux larger; it does not make the flux change. Therefore, however strong the magnets may be, no current can be generated in the stationary loop.

No current is induced in either case — neither when the loop is wholly inside the capacitor nor when it is partially outside.

An induced emf (and current) arises only from a changing magnetic flux through the circuit. The space between the capacitor plates contains an electric field, not a magnetic field. A changing electric flux does not drive a conduction current around a closed conducting loop.

Hence, in case (i) and in case (ii), there is no induced current. The induced current cannot be produced by changing the electric flux linked with the loop.

The induced emf is given by $$\varepsilon = B\dfrac{dA_{\text{in field}}}{dt}$$, where $$A_{\text{in field}}$$ is the area of the loop still inside the field region. Since $$B$$ and the speed $$v$$ are constant, the emf is constant only if the area leaves the field at a constant rate.

Rectangular loop: As it moves out at constant velocity, the length of the side cutting the field boundary stays the same, so the area inside the field decreases at a constant rate. Hence the induced emf is constant during the passage out.

Circular loop: As it moves out, the length of the chord at the field boundary keeps changing, so the area inside the field decreases at a non-uniform rate. Hence the induced emf is not constant — it varies during the passage out.

Therefore, the induced emf is expected to be constant only for the rectangular loop.

In Fig. 6.9, two bar magnets are moved towards the coil from opposite sides, changing the magnetic flux linked with the coil and inducing an emf. The ends of the coil are connected to the plates A and B of the capacitor.

Applying Lenz's law to the coil, the induced current charges the capacitor in such a way that plate A becomes positive with respect to plate B.

The energy stored in an inductor carrying a current $$I$$ is

$$U_B = \frac{1}{2}L I^2$$

For a solenoid of $$n$$ turns per unit length, length $$l$$ and cross-sectional area $$A$$, the self-inductance is

$$L = \mu_0 n^2 A l$$

and the magnetic field inside is $$B = \mu_0 n I$$, which gives

$$I = \frac{B}{\mu_0 n}$$

Substituting both into the energy expression:

$$U_B = \frac{1}{2}L I^2 = \frac{1}{2}\,(\mu_0 n^2 A l)\left(\frac{B}{\mu_0 n}\right)^2$$

$$U_B = \frac{1}{2}\,\mu_0 n^2 A l \times \frac{B^2}{\mu_0^2 n^2}$$

The factors $$n^2$$ cancel, leaving

$$U_B = \frac{B^2}{2\mu_0}\,A l$$

The magnetic energy per unit volume (energy density) of the solenoid is obtained by dividing $$U_B$$ by the volume $$V = Al$$ that contains the field:

$$u_B = \frac{U_B}{V} = \frac{U_B}{Al} = \frac{B^2}{2\mu_0}$$

The corresponding result for a parallel-plate capacitor — the electrostatic energy per unit volume (derived in Chapter 2) — is

$$u_E = \frac{1}{2}\varepsilon_0 E^2$$

Comparison: In both cases the energy density is proportional to the square of the field strength ($$B^2$$ for the magnetic field and $$E^2$$ for the electric field). Although these expressions were derived for the special cases of a solenoid and a parallel-plate capacitor respectively, they are in fact general and hold for any region of space containing a magnetic or an electric field.

By Lenz's law, the induced current opposes the change in flux that produces it.

In Fig. 6.15(a), the bar magnet is moving towards the coil pqr, so the magnetic flux linked with the coil is increasing. The induced current must flow so as to oppose this increase — i.e. the face of the coil towards the magnet must develop a polarity that repels the approaching magnet.

Applying this rule, the induced current in the coil flows along the path qrpq.

The bar magnet (S–N) moves along the common axis between the two coils. As it moves, the flux through one coil increases while that through the other decreases. By Lenz's law each induced current opposes its own change of flux.

For coil pqr, the magnet is approaching, so the flux through it increases; the induced current opposes the increase and flows along prq.

For coil xyz, the magnet is receding, so the flux through it decreases; the induced current opposes the decrease and flows along yzx.

When the tapping key of the first coil's circuit is just closed, the current in that coil grows from zero. This builds up a magnetic field, and the flux linked with the neighbouring (second) coil increases.

By Lenz's law, the induced current in the second coil opposes this increase in flux. The induced current therefore flows along yzx.

Changing the rheostat setting changes the current in the first coil, and hence the flux linked with the second coil. Taking the case in which the rheostat is adjusted so that the current in the first coil decreases, the flux through the second coil decreases.

By Lenz's law, the induced current in the second coil opposes this decrease in flux, and therefore flows along zyx.

When the tapping key of one solenoid is just released, its circuit is broken and the current in it falls rapidly to zero. The magnetic field it produced collapses, so the flux linked with the loop xyz decreases.

By Lenz's law, the induced current in the loop opposes this decrease in flux. The induced current therefore flows along xry.

The circular loop lies in the same plane as the straight current-carrying wire. The magnetic field produced by the straight wire forms concentric circles around the wire; at every point of this plane, these field lines lie in the plane of the loop itself.

Hence no magnetic field line passes through (normal to) the area of the loop — the magnetic flux linked with the loop is zero.

Even though the current $$I$$ is decreasing steadily, the flux through the loop remains zero throughout. Since there is no change of flux through the loop, no current is induced in it.

In Fig. 6.16(a), the magnetic field is directed into the plane of the page (shown by $$\times$$ marks). The wire of irregular shape is being turned into a circular shape.

For a given perimeter (a fixed length of wire), a circle encloses the maximum area. So as the irregular loop becomes circular, the area enclosed by it increases.

Since $$B$$ is into the page and the enclosed area increases, the magnetic flux into the page increases. By Lenz's law, the induced current must oppose this increase, i.e. it must produce a magnetic field out of the page inside the loop. This requires the induced current to flow anticlockwise, i.e. along the path adcb.

In Fig. 6.16(b), the magnetic field is directed out of the plane of the page (shown by dots). The circular loop is being deformed into a narrow straight wire.

For a given perimeter, a circle encloses the maximum area, while a narrow straight wire encloses almost zero area. So as the circular loop is deformed, the area enclosed by it decreases.

Since $$B$$ is out of the page and the enclosed area decreases, the magnetic flux out of the page decreases. By Lenz's law, the induced current must oppose this decrease, i.e. it must produce a magnetic field out of the page inside the loop. This requires the induced current to flow anticlockwise, i.e. along the path $$a'd'c'b'a'$$.

Given: rectangular loop with sides $$8\,\mathrm{cm} = 0.08\,\mathrm{m}$$ (longer) and $$2\,\mathrm{cm} = 0.02\,\mathrm{m}$$ (shorter); field $$B = 0.3\,\mathrm{T}$$ normal to the loop; speed $$v = 1\,\mathrm{cm\,s^{-1}} = 0.01\,\mathrm{m\,s^{-1}}$$.

(a) Velocity normal to the longer side. When the loop moves with its velocity perpendicular to the longer side, it is the longer side $$(l = 0.08\,\mathrm{m})$$ that cuts the field lines and acts as the source of motional emf:

$$\varepsilon = B\,l\,v = 0.3 \times 0.08 \times 0.01$$

$$\varepsilon = 2.4\times10^{-4}\,\mathrm{V}$$

Duration: the emf lasts only while the loop is partly inside the field. The loop crosses the field boundary over a distance equal to its dimension along the motion, which here is the shorter side $$0.02\,\mathrm{m}$$:

$$t = \frac{\text{distance}}{v} = \frac{0.02}{0.01} = 2\,\mathrm{s}$$

So the induced emf is $$2.4\times10^{-4}\,\mathrm{V}$$ and it lasts for $$2\,\mathrm{s}$$.

(b) Velocity normal to the shorter side. Now the velocity is perpendicular to the shorter side, so it is the shorter side $$(l = 0.02\,\mathrm{m})$$ that cuts the field lines and acts as the source of motional emf:

$$\varepsilon = B\,l\,v = 0.3 \times 0.02 \times 0.01$$

$$\varepsilon = 6.0\times10^{-5}\,\mathrm{V}$$

Duration: the loop now crosses the field boundary over a distance equal to its longer side $$0.08\,\mathrm{m}$$:

$$t = \frac{\text{distance}}{v} = \frac{0.08}{0.01} = 8\,\mathrm{s}$$

So the induced emf is $$6.0\times10^{-5}\,\mathrm{V}$$ and it lasts for $$8\,\mathrm{s}$$.

Note: The larger emf $$(2.4\times10^{-4}\,\mathrm{V})$$ in part (a) lasts for a shorter time $$(2\,\mathrm{s})$$, while the smaller emf $$(6.0\times10^{-5}\,\mathrm{V})$$ here lasts longer $$(8\,\mathrm{s})$$.

Given: length of wire $$l = 10\,\mathrm{m}$$, speed of fall $$v = 5.0\,\mathrm{m\,s^{-1}}$$, horizontal component of earth's field $$B = 0.30\times10^{-4}\,\mathrm{Wb\,m^{-2}}$$. The wire falls at right angles to $$B$$.

As the wire falls, it cuts the horizontal field lines, so a motional emf is induced:

$$\varepsilon = B\,l\,v$$

$$\varepsilon = (0.30\times10^{-4}) \times 10 \times 5.0$$

$$\varepsilon = 1.5\times10^{-3}\,\mathrm{V}$$

The instantaneous value of the induced emf is $$1.5\times10^{-3}\,\mathrm{V} = 1.5\,\mathrm{mV}$$.

The free charges in the wire experience the magnetic (Lorentz) force $$\mathbf{F} = q\,\mathbf{v}\times\mathbf{B}$$.

Take East as $$\hat{x}$$, North as $$\hat{y}$$ and the upward vertical as $$\hat{z}$$. The wire falls downward, so $$\mathbf{v} = -v\,\hat{z}$$. The horizontal component of the earth's field points North, so $$\mathbf{B} = B\,\hat{y}$$.

$$\mathbf{v}\times\mathbf{B} = (-v\,\hat{z})\times(B\,\hat{y}) = -vB\,(\hat{z}\times\hat{y}) = -vB\,(-\hat{x}) = vB\,\hat{x}$$

So the force on a positive charge is directed along $$+\hat{x}$$, i.e. towards the East. The induced emf therefore acts from the West end to the East end of the wire.

From part (b), the magnetic force pushes positive charges towards the East end of the wire. Positive charge therefore accumulates at the East end, leaving the West end negative.

Hence the East end of the wire is at the higher electrical potential.