# 15.2.3 Flux Change vs Flux Cut

Let’s recap our study of Faraday’s Laws so far.

Stationary Coil

First, there is the transformer emf: when a stationary coil of area A and N turns is facing a changing magnetic field B, the magnitude of the induced emf is proportional to the rate of change of magnetic flux linkage (of the coil). $\displaystyle \varepsilon =\frac{{d\Phi }}{{dt}}=\frac{{d(NBA)}}{{dt}}=NA\frac{{dB}}{{dt}}$

Notice that B has to vary at a constant rate to maintain a constant ε.

Moving Wire

Next, there is the motional emf: when a conductor of length L is moving at velocity v perpendicularly across a constant magnetic field B, the magnitude of the induced emf is proportional to the rate of flux cutting (by the moving conductor). $\displaystyle \varepsilon =BLv$

Notice that the conductor only has to maintain a constant v to maintain a constant ε.

Moving/Rotating/Deforming Coil

When a coil moves, or rotates, or deforms (e.g. shrinks or expands) in a constant magnetic field, emf may also be induced in the coil. This is fundamentally motional emf: a coil is made up of wire segments. Those wire segments are cutting magnetic flux as the coil moves/rotates/deforms. The magnitude of the induced emf in the coil can be obtained by summing up the motional emfs induced in all the segments. $\displaystyle \varepsilon =\sum{{BLv}}$

Alternatively, we can also see that as the coil moves/rotates/deforms, the magnetic flux in the coil is changing. Hence, the magnitude of the induced emf can also be deduced from $\displaystyle \varepsilon =\frac{{d\Phi }}{{dt}}=\frac{{d(BA)}}{{dt}}=B\frac{{dA}}{{dt}}$.

Both approaches are good, as long as you apply them correctly.

Example

A rectangular coil (30 cm by 20 cm) is moving at a constant speed of 5.0 cm s-1, entering and then leaving a region of uniform magnetic field of 0.80 T. The total resistance of the coil is 6.0 W. Determine the direction and magnitude of the induced current in the coil when it is
a) entering,
b) moving within, and
c) leaving the field.

Approach 1: Flux Cutting

a)

As the coil enters the field, only the PQ segment of the coil is cutting the flux (perpendicularly).

Using the FRHR, P and Q are the positive and negative terminals of the “battery” respectively. So the current is going to flow in an anticlockwise direction in the coil.

The emf induced between P and Q is therefore $\displaystyle \varepsilon =BLv=(0.80)(0.30)(0.050)=0.012\text{ V}$

The induced current is therefore $\displaystyle I=\frac{V}{R}=\frac{{0.012}}{{6.0}}=2.0\text{ mA}$

b)

When the entire coil is in the field, both the PQ and RS segments of the coil are cutting the flux. But note that PQ is trying to push current in the anticlockwise direction, whereas RS is trying to push current in the clockwise direction. The two emfs “cancel” each other. So the resultant emf and current is zero.

c)

As the coil leaves the field, only the RS segment of the coil is cutting the flux (perpendicularly).

Using the FRHR, R and S are the positive and negative terminals of the “battery” respectively. So the current is going to flow in a clockwise direction in the coil.

The magnitude of the emf and current is the same as part a).

Approach 2: Flux Changing

a) As the coil enters the field, a larger and larger area is exposed to the magnetic field. So the coil experiences an increasing magnetic flux (going into page). To oppose this change, the induced emf ought to be producing magnetic flux in the opposite direction (coming out of the page) (Lenz’s law). Using the RHGR, we can conclude that the induced current is going to be anticlockwise.

Since it takes $\displaystyle 20\div 5.0=4.0\text{ s}$ for the coil to complete the increase in flux, $\displaystyle \varepsilon =\frac{{d\phi }}{{dt}}=\frac{{d(BA)}}{{dt}}=\frac{{(0.80)(0.30\times 0.20)}}{{4.0}}=0.012\text{ V}$

The induced current is therefore $\displaystyle I=\frac{V}{R}=\frac{{0.012}}{{6.0}}=2.0\text{ mA}$

b) When the entire coil is in the field, the magnetic flux of the coil is constant. Immediately, we can conclude that there is no induced emf and current around the coil.

c)

As the coil leaves the field, a smaller and smaller area is exposed to the magnetic field. So it experiences a decreasing magnetic flux (going into the page). To oppose this change, the induced emf ought to be producing magnetic flux in the same direction (going into the page) (Lenz’s law). Using the RHGR, we can conclude that the induced current is going to be clockwise.

The magnitude of the emf is same as part a).

Concept Test

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