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`color{blue} ✍️` In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced a rule, known as Lenz’s law which gives the polarity of the induced emf in a clear and concise fashion. The statement of the law is:

`color{purple}{"he polarity of induced emf is such that it tends to produce a "}`
`color{purple}{"current which opposes the change in magnetic flux that produced it. "}`

`color{blue} ✍️`The negative sign shown in Eq. (6.3) represents this effect. We can understand Lenz’s law by examining Experiment 6.1 in Section 6.2.1. In Fig. 6.1, we see that the North-pole of a bar magnet is being pushed towards the closed coil.

`color{blue} ✍️`As the North-pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases.

`color{blue} ✍️`Hence current is induced in the coil in such a direction that it opposes the increase in flux. This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet.

`color{brown}bbul{"Note"}` that magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet.

`color{blue} ✍️`Similarly, if the Northpole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux, the induced current in the coil flows in clockwise direction and its South-pole faces the receding North-pole of the bar magnet.

`color{blue} ✍️`This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in flux.

`color{blue} ✍️`Suppose if an open circuit is used in place of the closed loop in the above example, In this case too, an emf is induced across the open ends of the circuit. The direction of the induced emf can be found using Lenz’s law.

`color{blue} ✍️`Consider Figs. 6.6 (a) and (b). They provide an easier way to understand the direction of induced currents. Note that the direction shown by indicate the directions of the induced currents.

`color{blue} ✍️`A little reflection on this matter should convince us on the correctness of Lenz’s law. Suppose that the induced current was in the direction opposite to the one depicted in Fig. 6.6(a).

`color{blue} ✍️` In that case, the South-pole due to the induced current will face the approaching North-pole of the magnet.
The bar magnet will then be attracted towards the coil at an ever increasing acceleration. A gentle push on the magnet will initiate the process and its velocity and kinetic energy will continuously increase without expending any energy.

`color{blue} ✍️`If this can happen, one could construct a perpetual-motion machine by a suitable arrangement. This violates the law of conservation of energy and hence can not happen.

`color{blue} ✍️`Now consider the correct case shown in Fig. 6.6(a). In this situation, the bar magnet experiences a repulsive force due to the induced current. Therefore, a person has to do work in moving the magnet.

`color{blue} ✍️`Where does the energy spent by the person go? This energy is dissipated by Joule heating produced by the induced current.
Q 3138545402

Figure 6.7 shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law.
Class 12 Chapter 6 Example 4

(i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field, The induced current must flow along the path bcdab so that it opposes the increasing flux.
(ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux.
(iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux. Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field.
Q 3158545404

(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets?
(b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop.
(c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Fig. 6.8) to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops.
Class 12 Chapter 6 Example 5

(a) No. However strong the magnet may be, current can be induced
only by changing the magnetic flux through the loop.
(b) No current is induced in either case. Current can not be induced by changing the electric flux.
(c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.
(d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.


`color{blue} ✍️`Let us consider a straight conductor moving in a uniform and timeindependent magnetic field. Figure 6.10 shows a rectangular conductor PQRS in which the conductor PQ is free to move.

`color{blue} ✍️`The rod `PQ` is moved towards the left with a constant velocity `v` as shown in the figure. Assume that there is no loss of energy due to friction. `PQRS` forms a closed circuit enclosing an area that changes as `PQ` moves. It is placed in a uniform magnetic field `B` which is perpendicular to the plane of this system.

`color {blue}{➢➢}` If the length `RQ = x` and `RS = l,` the magnetic flux `ΦB` enclosed by the loop `PQRS` will be

`color{fuchsia}(ΦB = Blx)`

`color {blue}{➢➢}`Since `x` is changing with time, the rate of change of flux `ΦB` will induce an emf given by:

`color{purple}(ε= (-dΦ_B)/(dt) = - d/(dt) (Blx))`

`color{purple}(= - Bl(dx)/(dt) = Blv)`


`color {blue}{➢➢}`where we have used `dx//dt = –v` which is the speed of the conductor `PQ`.

`color{blue} ✍️`The induced emf `Blv` is called motional emf. Thus, we are able to produce induced emf by moving a conductor instead of varying the magnetic field, that is, by changing the magnetic flux enclosed by the circuit.

`color{blue} ✍️`It is also possible to explain the motional emf expression in Eq. (6.5) by invoking the Lorentz force acting on the free charge carriers of conductor PQ. Consider any arbitrary charge q in the conductor `PQ`.

`color{blue} ✍️`When the rod moves with speed v, the charge will also be moving with speed `v` in the magnetic field `B`. The Lorentz force on this charge is `qvB` in magnitude, and its direction is towards `Q`. All charges experience the same force, in magnitude and direction, irrespective of their position in the rod `PQ`. The work done in moving the charge from `P` to `Q` is,

`color{purple}(W = qvBl)`

`color {blue}{➢➢}`Since emf is the work done per unit charge,


`color{blue} ✍️`This equation gives emf induced across the rod `PQ` and is identical to Eq. (6.5). We stress that our presentation is not wholly rigorous. But it does help us to understand the basis of Faraday’s law when the conductor is moving in a uniform and time-independent magnetic field.

`color{blue} ✍️`On the other hand, it is not obvious how an emf is induced when a conductor is stationary and the magnetic field is changing – a fact which Faraday verified by numerous experiments. In the case of a stationary conductor, the force on its charges is given by

`color{purple}(F = q (E + v × B) = qE)`


`color {blue}{➢➢}` since `v = 0.` Thus, any force on the charge must arise from the electric field term `E` alone.

`color{blue} ✍️`Therefore, to explain the existence of induced emf or induced current, we must assume that a time-varying magnetic field generates an electric field.

`color{blue} ✍️`However, we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields.

`color{blue} ✍️`As we learnt that charges in motion (current) can exert force/torque on a stationary magnet. Conversely, a bar magnet in motion (or more generally, a changing magnetic field) can exert a force on the stationary charge. This is the fundamental significance of the Faraday’s discovery. Electricity and magnetism are related.
Q 3178545406

A metallic rod of 1 m length is rotated with a frequency of `50 rev//s,` with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (Fig. 6.11). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?
Class 12 Chapter 6 Example 6

Method I
As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. Using Eq. (6.5), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by

dε = Bv dr . Hence

`ε = int dε Bv dr = ε_(0)^(R) Bv dr= int_(0)^(R) b omegar dr = (BomegaR^2)/2`

Note that we have used `v = ω r.` This gives

`ε = 1/2 xx 1.0 xx 2pi xx50 xx(1^2)`

Method II
To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by

`pi R^2 xxtheta/(2pi) = 1/2 R^2theta`
where R is the radius of the circle. Hence, the induced emf is
`ε = Bxxd/(dt) [1/2 R^2 theta] = 1/2 BR^2 (dtheta)/(dt) = (BomegaR^2)/2`

`[Note: (dtheta)/(dt) = omega= 2pi v]`

This expression is identical to the expression obtained by Method I and we get the same value of ε.
Q 3108545408

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field HE at a place. If `HE = 0.4` G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1 G = 10^(–4) T.
Class 12 Chapter 6 Example 7

Induced emf `= (1/2) ω B R^2`

`= (1//2) × 4π × 0.4 × 10^(–4) × (0.5)^2`
`= 6.28 × 10^(–5) V`

The number of spokes is immaterial because the emf’s across the spokes are in parallel.