Faraday's Law of Electromagnetic Induction & Lenz's Law - Physics Class

Magnetic Flux

Consider a surface that is bounded by a closed curve and has area A. Let there be a (approximately) uniform magnetic field B in the region of the surface. The magnetic flux through the surface is

φ = B A cos θ

where θ is the angle between the magnetic field vector B and the unit normal to the surface (the area vector). If the magnetic field is parallel to the surface (i.e. perpendicular to the normal), then cos θ = 0 and the flux through the surface is zero. The SI unit of magnetic flux is the weber (Wb).

Notes :

Area vector is perpendicular to the surface.

For an open surface choose one direction as the area-vector direction and stick to it throughout the problem.

For closed surfaces the outward normal is conventionally taken as the area-vector direction.

Flux is basically the count of magnetic field lines crossing a surface (weighted by angle and field strength).

Magnetic field lines form closed loops; there are no magnetic monopoles (field lines do not begin or end).

Faraday's Law of Electromagnetic Induction

When the magnetic flux φ through the area bounded by a closed conducting loop changes with time, an electromotive force (emf) is induced around the loop. Faraday's law states:

ε = - dφ_B/dt

The negative sign gives the sense of the induced emf (see Lenz's law) - it indicates that the induced emf acts so as to oppose the change in flux.

The magnetic flux is given by

• φ_B = B · A · cos θ, where B is the magnetic-field magnitude, A is the area magnitude and θ is the angle between B and the area normal.

If the conducting loop has resistance R, the induced emf drives a current I given by Ohm's law:

I = ε / R

Combining with Faraday's law:

I = -(1/R) · dφ_B/dt

Faraday's law together with Ohm's law shows how a changing magnetic flux produces an induced current in a closed conductor.

Simple illustrative examples

Ex. A coil is placed in a constant magnetic field. The magnetic field is parallel to the plane of the coil as shown in figure. Find the emf induced in the coil.

Sol.

Since the magnetic field is parallel to the plane of the coil, the field is perpendicular to the area normal of the coil.

The flux φ = B A cos 90° = 0 for all times, so dφ/dt = 0 and hence ε = 0.

Therefore, the induced emf is zero.

Ex. Find the emf induced in the coil shown in figure. The magnetic field is perpendicular to the plane of the coil and is constant.

Sol.

When the magnetic field is perpendicular to the plane of the coil, the flux φ = B A cos 0° = B A, which is constant in time.

Therefore, dφ/dt = 0 and ε = 0.

Lenz's Law

Lenz's law states: The direction of the induced emf (and the induced current in a closed conductor) is such that it opposes the change in magnetic flux that produced it.

Lenz's law is a consequence of the conservation of energy: the induced current produces magnetic effects that oppose the change in flux, and work must be done against this opposition to change the flux.

    (a)                (c)           (b)            (d)

In figures (a) and (b), as the magnet approaches the loop the magnetic flux through the loop increases. The induced current produces its own magnetic field B_ind whose flux opposes the increase; hence B_ind is opposite in direction to the external field B_ext.

In figures (c) and (d), as the magnet moves away the flux through the loop decreases. The induced field tries to maintain the original flux, so B_ind is in the same direction as B_ext.

Because the induced current opposes the change, a force opposing the motion of the magnet appears. To move the magnet at constant velocity an external agent must do work; that mechanical work is converted into electrical energy in the circuit.

There are alternative practical methods to find the direction of induced current. One is to evaluate the sign of dφ/dt and then use Faraday's law; another is to use the right-hand rule and Lenz's law together to determine the sense of the induced current.

Consider a conducting loop placed near a long straight wire carrying current i. If the wire current increases, the magnetic field at the loop increases. Using the right-hand rule we find the directions of the field and the loop normal, determine the sign of the flux and its time derivative, then use Faraday's law and deduce the direction of induced current.

Brain teaser

Two identical coaxial circular loops carry equal currents circulating in the same direction. What will happen to the current in each loop if the loops approach each other?

When the loops approach, the mutual flux through each loop increases. By Lenz's law the induced currents oppose the change: each loop will experience an induced emf that opposes the increase of the current producing the approaching loop. The induced emf acts to reduce the net increase; if both currents are externally maintained, induced currents will appear that oppose the external change.

Ex. Find the direction of induced current in the coil shown in figure. Magnetic field is perpendicular to the plane of coil and it is increasing with time.

Sol.

Inward flux is increasing with time. To oppose the increase, the induced magnetic field must be outward. Such an outward induced field corresponds to an anticlockwise induced current (as viewed in the figure).

Ex. Figure shows a long current-carrying wire and two rectangular loops moving with velocity v. Find the direction of current in each loop.

Sol.

In loop (i) the magnetic flux through the loop does not change with time as it moves; therefore no emf is induced.

In loop (ii) the loop moves into a region of decreasing inward magnetic field. The flux (inward) is decreasing, so to oppose the decrease the induced field must be inward. To produce an inward induced field the induced current must be clockwise (as shown).

Calculation of Induced EMF

Magnetic flux linked with a closed conducting loop is φ = B A cos θ. A change in any one of B, A or θ (the angle between B and the area normal) will change φ and hence produce an induced emf.

By changing the magnetic field

Ex. Figure shows a coil placed in decreasing magnetic field applied perpendicular to the plane of the coil. The magnetic field is decreasing at a rate of 10 T/s. Find out current in magnitude and direction.

Sol.

The flux linked with the coil is φ = B A.

Emf magnitude |ε| = A · |dB/dt|.

Given A = 2 m² and dB/dt = -10 T/s (decreasing), so |ε| = 2 × 10 = 20 V.

Given resistance R = 5 Ω, current magnitude i = ε/R = 20/5 = 4 A.

Direction: Since B (into or out of plane as per figure) is decreasing, the induced current will try to oppose the decrease. From Lenz's law the induced current is anticlockwise.

Ex. The magnetic flux (φ₂) in a closed circuit of resistance 20 Ω varies with time (t) according to the equation φ = 7t² - 4t where φ is in weber and t is in seconds. The magnitude of the induced current at t = 0.25 s is

(A) 25 mA (B) 0.025 mA (C) 47 mA (D) 175 mA

Sol.

φ = 7t² - 4t.

ε = - dφ/dt = - (14t - 4).

At t = 0.25 s, ε = - (14 × 0.25 - 4) = - (3.5 - 4) = +0.5 V (magnitude 0.5 V).

Current magnitude i = |ε|/R = 0.5 / 20 = 0.025 A = 25 mA.

Therefore answer: (A) 25 mA.

Ex. Consider a long infinite wire carrying a time-varying current i = k t (k > 0). A circular loop of radius a and resistance R is placed with its centre at a distance d from the wire (a ≪ d). Find the induced current in the loop.

Sol. At distance r=d from a long straight wire,

Since a≪d, the magnetic field over the loop can be taken as uniform and equal to its value at the centre.

Area of the loop: $A\; =\; \backslash pi\; a^2$A = πa²

^{Magnetic flux:} 

^{Substitute $i\; =\; kt$i=kt:}

^{Induced emf (Faraday's law)}

^{Using Ohm's law,}

Ex. Two concentric coplanar circular loops have diameters 20 cm and 2 m and resistance of unit length of the wire = 10⁻⁴ Ω/m. A time-dependent voltage V = (4 + 2.5 t) volts is applied to the larger as shown. The current in the smaller loop is

Sol.

Resistance of larger loop

Current in the larger loop

Magnetic field at the centre due to larger loop:

Area of smaller loop:

Substituting values:

Induced emf in smaller loop

Resistance of smaller loop

Induced current in smaller loop

$B\; =\; \backslash frac\{\backslash mu\_0\; I\_2\}\{2r\_2\}$$I\_2(t)\; =\; \backslash frac\{V(t)\}\{R\_2\}\; =\; \backslash frac\{4\; +\; 2.5t\}\{2\backslash pi\; \backslash times\; 10^\{-4\}\}$Ex. A rectangular wire frame of length 0.2 m is located at a distance of 5 × 10⁻² m from a long straight wire carrying a current of 10 A as shown in the figure. The width of the frame = 0.05 m. The wire is in the plane of the rectangle. Find the magnetic flux through the rectangular circuit. If the current decays uniformly to 0 in 0.2 s, find the emf induced in the circuit.

Sol. 

Magnetic field at distance $r$r from a long straight wire:

Consider an elemental strip of width $dr$dr at distance $r$r.

Area of strip: dA = l dr

Flux through the strip:

Total magnetic flux:

Substitute values:

 Induced emf in the circuit

Current decreases uniformly from $10A10\backslash \; \backslash text\{A\}$ A to 0 in $s0.2\backslash \; \backslash text\{s\}$0.2 s:

Induced emf:

Substitute values:

$B\; =\; \backslash frac\{\backslash mu\_0\; I\}\{2\backslash pi\; r\}$

Ex. Figure shows a wire frame PQSTXYZ placed in a time-varying magnetic field given as B = b t, where b is a positive constant. Resistance per unit length of the wire is l. Find the current induced in the wire and draw its electrical equivalent diagram.

Sol. Magnetic flux through one rectangular loop:

Induced emf in one loop:

Magnitude of induced emf in each loop: E = Ab

Since B = bt increases with time, the induced current in both loops will try to oppose the increase.

From the figure:

• Induced emfs in the upper and lower loops are in opposite directions along the common segment at $T$T.
• Hence, the emfs aid each other in driving current through the frame.

So, net emf around the circuit: E_net = Ab + Ab = 2Ab

Resistance per unit length $=\; \backslash lambda$= λ

Total resistance of both rectangular loops:

$R\; =\; \backslash lambda\; \backslash times\; (\backslash text\{total\; wire\; length\})$R = λ × (total wire length)

If each loop has perimeter $L$L, then:

 R = 2 λ L

Using Ohm's law:

Brain teaser

A copper ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. Will the acceleration of the falling magnet be equal to, greater than or lesser than the acceleration due to gravity?

As the magnet approaches and passes through the ring, flux through the ring changes; eddy/induced currents in the ring produce magnetic fields that oppose the motion of the magnet (Lenz's law). Thus the magnet experiences a magnetic braking force and its acceleration will be less than g while it interacts with the ring.

By changing the area

Changing the area A of a loop (or changing its orientation so that θ changes) while the magnetic field is present also changes flux and induces emf. The same Faraday-Lenz relation applies: ε = - dφ/dt = - d(B A cos θ)/dt.

Solved Examples :

Ex. A space is divided by the line AD into two regions. Region I is field free and the region II has a uniform magnetic field B directed into the paper. ACD is a semicircular conducting loop of radius r with centre at O, the plane of the loop being in the plane of the paper. The loop is now made to rotate with a constant angular velocity ω about an axis passing through O, and perpendicular to the plane of the paper in the clockwise direction. The effective resistance of the loop is R.

(a) Obtain an expression for the magnitude of the induced current in the loop.

(b) Show the direction of the current when the loop is entering into the region II.

(c) Plot a graph between the induced emf and the time of rotation for two periods of rotation.

Sol. (a) When the loop is rotated about an axis passing through center O and perpendicular to the plane of the paper, the angle between magnetic field vector B and area A is always $0^\backslash cir$0^∘. When the loop $BA\; \backslash cos\; 0\; =\; 0$BA cos0 = 0 the (Since $B\; =\; 0$B = 0 in region I).

When the loop enters the magnetic field in region II, the magnetic flux linked with it is given by

Therefore, emf induced

As resistance of the loop is R, the current induced is given by

This is the required expression for current induced in the loop.

(b) According to Lenz's law, the direction of current induced is to oppose the change in magnetic flux. So, when entering into region II the field produced by the current induced anticlockwise as shown in Fig.

(c) When the loop enters the magnetic flux linked with it increases and the emf

is induced in one direction. when the loop comes out of the field the flux decreases and emf is induced in opposite sence. The graph for representing the emf induced versus time for two taken aniclockwise direction as positive.

Ex. Two parallel, long, straight conductors lie on a smooth plane surface. Two other parallel conductors rest on them at right angles so as to form a square of side a initially. A uniform magnetic field B exists at right angles to the plane containing the conductors. Now they start moving out with a constant velocity v. (a) Will the induced emf be time dependent? (b) Will the current be time dependent?

Sol. (a) As the conductors move outward with constant velocity v, the side length of the square increases uniformly with time.

Let at time t, the side of the square be:

Area of the square:

Induced emf:

Since emf depends on time t, the induced emf is time dependent.

(b) The resistance of the square loop depends on its total length, which increases as the conductors move outward.

At time $t$t, 

If resistance per unit length is constant, then:

Current induced:

Since,