MAGNETIC FLUX
Magnetic Flux Definition
- Electromagnetic induction is when an e.m.f is induced in a closed circuit conductor due to it moving through a magnetic field
- This happens when a conductor cuts through magnetic field lines
- The amount of e.m.f induced is determined by the magnetic flux
- This is the total magnetic field that passes through a given area
- It is a maximum when the magnetic field lines are perpendicular to the area
- It is at a minimum when the magnetic field lines are parallel to the area
- The magnetic flux is defined as:
The product of the magnetic flux density and the cross-sectional area perpendicular to the direction of the magnetic flux density - In other words, magnetic flux is the number of magnetic field lines through a given area
Calculating Magnetic Flux
- Magnetic flux is defined by the symbol $\Phi$ (greek letter ‘phi’)
- It is measured in units of Webers (Wb)
- Magnetic flux can be calculated using the equation:
$
\boldsymbol{\Phi}=\mathbf{B A}
$
- Where:
- $\Phi=$ magnetic flux $(W b)$
- $B=$ magnetic flux density $(T)$
- $A=$ cross-sectional area $\left(\mathrm{m}^2\right)$
- When the magnet field lines are not completely perpendicular to the area $A$, then the component of magnetic flux density $B$ perpendicular to the area is taken
- The equation then becomes:
$
\Phi=B A \cos (\theta)
$
- Where:
- $\theta=$ angle between magnetic field lines and the line perpendicular to the plane of the area (often called the normal line) (degrees)
- This means the magnetic flux is:
- Maximum $=B A$ when $\cos (\theta)=1$ therefore $\boldsymbol{\theta}=0^{\circ}$. The magnetic field lines are perpendicular to the plane of the area
- Minimum $=0$ when $\cos (\theta)=0$ therefore $\boldsymbol{\theta}=\mathbf{9 0}$. The magnetic fields lines are parallel to the plane of the area
- An e.m.f is induced in a circuit when the magnetic flux linkage changes with respect to time
- This means an e.m.f is induced when there is:
- A changing magnetic flux density $B$
- A changing cross-sectional area $A$
- A change in angle $\theta$
Worked example
An aluminium window frame has a width of $40 \mathrm{~cm}$ and length of $73 \mathrm{~cm}$ as shown in the diagram below.
The frame is hinged along the vertical edge AC.
When the window is closed, the frame is normal to the Earth’s magnetic field with magnetic flux density $1.8 \times 10^{-5} \mathrm{~T}$.
a) Calculate the magnetic flux through the window when it is closed.
b) Sketch the graph of the magnetic flux against angle between the field lines and the normal when the window is opened and rotated by $180^{\circ}$.
Answer/Explanation
Part (a)
Step 1: Write out the known quantities
Cross-sectional area, $A=40 \mathrm{~cm} \times 73 \mathrm{~cm}=\left(40 \times 10^{-2}\right) \times\left(73 \times 10^{-2}\right)=0.292$ $m^2$
Magnetic flux density, $B=1.8 \times 10^{-5} \mathrm{~T}$
Step 2: Write down the equation for magnetic flux
$
\Phi=\mathbf{B A}
$
Step 3:
Substitute in values
$
\Phi=\left(1.8 \times 10^{-5}\right) \times 0.292=5.256 \times 10^{-6}=5.3 \times 10^{-6} \mathrm{~Wb}
$
Part (b)
The magnetic flux will be at a minimum when the window is opened by $90^{\circ}$ and a maximum when fully closed or opened to $180^{\circ}$
Exam Tip
Consider carefully the value of $\theta$, it is the angle between the field lines and the line normal (perpendicular) to the plane of the area the field lines are passing through. If it helps, drawing the normal on the area provided will help visualise the correct angle.
MAGNETIC FLUX LINKAGE
Magnetic Flux Linkage
- The magnetic flux linkage is a quantity commonly used for solenoids which are made of $N$ turns of wire
- Magnetic flux linkage is defined as:
- The product of the magnetic flux and the number of turns
- It is calculated using the equation:
$
\Phi \mathbf{N}=\mathbf{B A N}
$
- Where:
- $\Phi=$ magnetic flux $(W b)$
- $\mathrm{N}=$ number of turns of the coil
- $B=$ magnetic flux density $(T)$
- $\mathrm{A}=$ cross-sectional area $\left(\mathrm{m}^2\right)$
- The flux linkage $\Phi N$ has the units of Weber turns (Wb turns)
- As with magnetic flux, if the field lines are not completely perpendicular to the plane of the area they are passing through
- Therefore, the component of the flux density which is perpendicular is equal to:
$
\Phi N=B A N \cos (\theta)
$
Worked example
(2) A solenoid of circular cross-sectional radius $0.40 \mathrm{~m}^2$ and 300 turns is placed perpendicular to a magnetic field with a magnetic flux density of $5.1 \mathrm{mT}$. Determine the magnetic flux linkage for this solenoid.
Answer/Explanation
Step 1:
Write out the known quantities
Cross-sectional area, $A=\pi r^2=\pi(0.4)^2=0.503 \mathrm{~m}^2$
Magnetic flux density, $B=5.1 \mathrm{mT}$
Number of turns of the coil, $\mathrm{N}=\mathbf{3 0 0}$ turns
Step 2:
Write down the equation for the magnetic flux linkage
$
\Phi N=B A N
$
Step 3:
Substitute in values and calculate
$
\Phi N=\left(5.1 \times 10^{-3}\right) \times 0.503 \times 300=0.7691=0.8 \mathrm{~Wb} \text { turns }(2 \mathrm{s.f})
$
PRINCIPLES OF ELECTROMAGNETIC INDUCTION
Principles of Electromagnetic Induction
- Electromagnetic induction is a phenomenon which occurs when an e.m.f is induced when a conductor moves through a magnetic field
- When the conductor cuts through the magnetic field lines:
- This causes a change in magnetic flux
- Which causes work to be done
- This work is then transformed into electrical energy
- Therefore, if attached to a complete circuit, a current will be induced
- This is known as electromagnetic induction and is defined as:
- The process in which an e.m.f is induced in a closed circuit due to changes in magnetic flux
- This can occur either when:
- A conductor cuts through a magnetic field
- The direction of a magnetic field through a coil changes
- Electromagnetic induction is used in:
- Electrical generators which convert mechanical energy to electrical energy
- Transformers which are used in electrical power transmission
- This phenomenon can easily be demonstrated with a magnet and a coil, or a wire and two magnets
Experiment 1: Moving a magnet through a coil
- When a coil is connected to a sensitive voltmeter, a bar magnet can be moved in and out of the coil to induce an e.m.f
The expected results are:
- When the bar magnet is not moving, the voltmeter shows a zero reading
- When the bar magnet is held still inside, or outside, the coil, the rate of change of flux is zero, so, there is no e.m.f induced
- When the bar magnet begins to move inside the coil, there is a reading on the voltmeter
- As the bar magnet moves, its magnetic field lines ‘cut through’ the coil, generating a change in magnetic flux
- This induces an e.m.f within the coil, shown momentarily by the reading on the voltmeter
- When the bar magnet is taken back out of the coil, an e.m.f is induced in the opposite direction
- As the magnet changes direction, the direction of the current changes
- The voltmeter will momentarily show a reading with the opposite sign
- Increasing the speed of the magnet induces an e.m.f with a higher magnitude
- As the speed of the magnet increases, the rate of change of flux increases
- The direction of the electric current, and e.m.f, induced in the conductor is such that it opposes the change that produces it
- Factors that will increase the induced e.m.f are:
- Moving the magnet faster through the coil
- Adding more turns to the coil
- Increasing the strength of the bar magnet
- Experiment 2: Moving a wire through a magnetic field
- When a long wire is connected to a voltmeter and moved between two magnets, an e.m.f is induced
- Note: there is no current flowing through the wire to start with
The expected results are:
- When the wire is not moving, the voltmeter shows a zero reading
- When the wire is held still inside, or outside, the magnets, the rate of change of flux is zero, so, there is no e.m.f induced
- As the wire is moved through between the magnets, an e.m.f is induced within the wire, shown momentarily by the reading on the voltmeter
- As the wire moves, it ‘cuts through’ the magnetic field lines of the magnetic, generating a change in magnetic flux
- When the wire is taken back out of the magnet, an e.m.f is induced in the opposite direction
- As the wire changes direction, the direction of the current changes
- The voltmeter will momentarily show a reading with the opposite sign
- As before, the direction of the electric current, and e.m.f, induced in the conductor is such that it opposes the change that produces it
- Factors that will increase the induced e.m.f are:
- Increasing the length of the wire
- Moving the wire between the magnets faster
- Increasing the strength of the magnets
FARADAY’S & LENZ’S LAWS
Faraday’s & Lenz’s Laws
- Faraday’s law tells us the magnitude of the induced e.m.f in electromagnetic induction and is defined as:
The magnitude of the induced e.m.f is directly proportional to the rate of change in magnetic flux linkage
$
\varepsilon=N \frac{\Delta \phi}{\Delta t}
$
- Where:
$
\begin{aligned}
& -\varepsilon=\text { induced e.m.f }(\mathrm{V}) \\
& -\mathrm{N}=\text { number of turns of coil } \\
& -\Delta \phi=\text { change in magnetic flux (Wb) } \\
& -\Delta t=\text { time interval (s) }
\end{aligned}
$
- Lenz’s Law gives the direction of the induced e.m.f as defined by Faraday’s law:
The induced e.m.f acts in such a direction to produce effects which oppose the change causing it - Lenz’s law combined with Faraday’s law is:
$
\varepsilon=-\mathrm{N} \frac{\Delta \phi}{\Delta t}
$
- This equation shows:
- When a bar magnet goes through a coil, an e.m.f is induced within the coil due to a change in magnetic flux
- A current is also induced which means the coil now has its own magnetic field
- The coil’s magnetic field acts in the opposite direction to the magnetic field of the bar magnet
- If a direct current (d.c) power supply is replaced with an alternating current (a.c) supply, the e.m.f induced will also be alternating with the same frequency as the supply
Experimental Evidence for Lenz’s Law
- To verify Lenz’s law, the only apparatus needed is:
- A bar magnet
- A coil of wire
- A sensitive ammeter
- Note: a cell is not required
- A known pole (either north or south) of the bar magnet is pushed into the coil, which induces a magnetic field in the coil
- Using the right hand grip rule, the curled fingers indicate the direction of the current and the thumb indicates the direction of the induced magnetic field
- The direction of the current is observed on the ammeter
- Reversing the magnet direction would give an opposite deflection on the meter
- The induced field (in the coil) repels the bar magnet
- This is because of Lenz’s law:
- The direction of the induced field in the coil pushes against the change creating it, ie. the bar magnet
Worked example: Faraday’s & Lenz’s Laws
A small rectangular coil contains 350 turns of wire. The longer sides are $3.5 \mathrm{~cm}$ and the shorter sides are $1.4 \mathrm{~cm}$.
The coil is held between the poles of a large magnet so that the coil can rotate about an axis through its centre.
The magnet produces a uniform magnetic field of flux density $80 \mathrm{mT}$ between its poles. The coil is positioned horizontally and then turned through an angle of $40^{\circ}$ in a time of $0.18 \mathrm{~s}$.
Calculate the magnitude of the average e.m.f induced in the coil.
Answer/Explanation
Step 1:
Write down the known quantities
Magnetic flux density, $B=80 \mathrm{mT}=80 \times 10^{-3} \mathrm{~T}$
Area, $A=3.5 \times 1.4=\left(3.5 \times 10^{-2}\right) \times\left(1.4 \times 10^{-2}\right)=4.9 \times 10^{-4} \mathrm{~m}^2$
Number of turns, $N=350$
Time interval, $\Delta t=0.18 \mathrm{~s}$
Angle between coil and field lines, $\theta=40^{\circ}$
Step 2: Write out the equation for Faraday’s law:
$
\varepsilon=\mathrm{N} \frac{\Delta \phi}{\Delta t}
$
Step 3:
Write out the equation for flux linkage:
$
\phi N=B A N \cos (\theta)
$
Step 4: Substitute values into flux linkage equation:
$
\phi N=\left(80 \times 10^{-3}\right) \times\left(4.9 \times 10^{-4}\right) \times 350 \times \cos (40)=0.0105 \mathrm{~Wb} \text { turns }
$
Step 5:
Substitute flux linkage and time into Faraday’s law equation:
$
\varepsilon=\frac{0.0105}{0.18}=0.05839=58 \mathrm{mV}(2 \text { s.f. })
$