10 7 Lenzs Law And Direction Of Induced Emf – Class 11TH Physics

10.7 Lenz's Law and Direction of Induced EMF

In the previous section, a mathematical expression for Faraday's law of electromagnetic induction was derived as:

\varepsilon = -N\left(\frac{\Delta\phi}{\Delta t}\right)

The minus sign in the expression is very important; it relates to the direction of the induced emf. To determine the direction, we use a principle based on the discovery made by the Russian physicist Heinrich Lenz in 1834. He found that the polarity of an induced emf always produces an induced current that opposes the change in the magnetic field that causes the emf. This principle is known as Lenz's Law, which states that:

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The direction of the induced current is always such that it opposes the change that causes the current.

Lenz's Law specifically applies to induced currents and not directly to induced emf. This means we can apply Lenz's Law to closed conducting loops or coils. If the loop is not closed, we can imagine it as if it were closed to determine the direction of the induced current, and from this, we can infer the direction of the induced emf.

Let us apply Lenz's law to a coil in which a current is induced by the movement of a bar magnet. A current-carrying coil generates a magnetic field similar to that of a bar magnet, with one face of the coil acting as the north pole and the other as the south pole. To oppose the motion of the bar magnet, the face of the coil towards the magnet must become a north pole (Fig. 10.15). This arrangement causes the two north poles to repel each other. According to the right-hand rule, the induced current in the coil must flow anticlockwise when viewed from the side of the bar magnet.

According to Lenz's law, the "push" of the magnet is the "change" that produces the induced current, and the current acts to oppose the push. On the other hand, if we pull the magnet away from the coil, the induced current will oppose the "pull" by creating a south pole on the face of the coil towards the bar magnet.

Lenz's law is also a manifestation of the law of conservation of energy and can be conveniently applied to circuits involving induced currents. To understand this, let us revisit the experiment depicted in Fig. 10.16. When the rod moves to the right, an emf is induced in it, causing an induced current to flow through the loop in anticlockwise direction. Because the current-carrying rod is moving within the magnetic field, it experiences a magnetic force $F_{m}$ with the magnitude of $F_{m} = ILB\sin\theta$ .

According to the right-hand rule, the direction of the magnetic force $F_{m}$ is opposite to that of the velocity $v$ , so it tends to stop the rod (Fig. 10.16-a). To keep the rod moving with a constant velocity, an external force equal in magnitude to $F_{m}$ but opposite in direction must be applied. This external force provides the energy necessary for the induced current to flow. Thus, electromagnetic induction adheres to the law of conservation of energy.

The Lenz's law forbids the induced current directed clockwise in this case, because the force $F_{m}$ would be, then, in the direction of $v$ that would accelerate the rod towards right (Fig. 10.16-b). This in turn would induce a stronger current, the magnetic field due to it also increases and the magnetic force increases further. Thus, the motion of the wire is accelerated more and more. Starting with a minute quantity of energy, we obtain an ever-increasing kinetic energy of motion apparently from nowhere. Consequently, the process becomes self-perpetuating which is against the law of conservation of energy.

Example 10.6

A metal rod of length $25\,\mathrm{cm}$ is moving at a speed of $0.5\,\mathrm{m\,s^{-1}}$ in a direction perpendicular to a $0.25\,\mathrm{T}$ magnetic field. Find the emf produced in the rod.

Solution

Speed of rod $v = 0.5\,\mathrm{m\,s^{-1}}$

Length of rod $L = 25\,\mathrm{cm} = 0.25\,\mathrm{m}$

Magnetic flux density $B = 0.25\,\mathrm{T} = 0.25\,\mathrm{N\,A^{-1}\,m^{-1}}$

Induced emf $\varepsilon = ?$

Using the relation: $\varepsilon = vBL$

\varepsilon = 0.5\,\mathrm{m\,s^{-1}} \times 0.25\,\mathrm{N\,A^{-1}\,m^{-1}} \times 0.25\,\mathrm{m}

\varepsilon = 3.13 \times 10^{-2}\,\mathrm{J\,C^{-1}} = 3.13 \times 10^{-2}\,\mathrm{V}

Example 10.7

A loop of wire is placed in a uniform magnetic field that is perpendicular to the plane of the loop. The strength of the magnetic field is $0.6\,\mathrm{T}$ . The area of the loop begins to shrink at a constant rate of $\Delta A/\Delta t = 0.8\,\mathrm{m^{2}\,s^{-1}}$ . What is the magnitude of emf induced in the loop while it is shrinking?

Solution

Rate of change of area $= \Delta A/\Delta t = 0.8\,\mathrm{m^{2}\,s^{-1}}$

Magnetic flux density $= B = 0.6\,\mathrm{T} = 0.6\,\mathrm{N\,A^{-1}\,m^{-1}}$

Number of turns $= N = 1$

Induced emf $= \varepsilon = ?$

Rate of change of flux $= \Delta\phi/\Delta t = B\cdot\Delta A/\Delta t$

Applying Faraday's law, magnitude of induced emf is:

\varepsilon = N\left(\frac{\Delta\phi}{\Delta t}\right) = NB\left(\frac{\Delta A}{\Delta t}\right)

= 1 \times 0.6\,\mathrm{N\,A^{-1}\,m^{-1}} \times 0.8\,\mathrm{m^{2}\,s^{-1}} = 0.48\,\mathrm{J/C} = 0.48\,\mathrm{V}