Motional emf

A rod in a field

So far the flux changed because the field changed. Now let the field stay steady and move a conductor in it. An emf still appears. It is called a motional emf.

Take a metal rod of length $l$. Move it with a speed $v$ at right angles to its length, in a uniform field $B$ that is at right angles to both the rod and the velocity. The free electrons in the rod move with it, so a magnetic force acts on each one. It is the force on a moving charge $q$ that you know, $F = qvB$.

The force pushes the electrons along the rod to one end. That end becomes negative, and the other end is left positive. The separated charges make an electric field $E$ inside the rod, and it pushes back on the electrons. The pile-up stops when the two forces balance:

The potential difference between the ends of the rod is $E\,l$. So the rod behaves like a battery of emf

The speed, the length and the field must be at right angles to one another.

The same result from Faraday's law

Put the rod across two rails that are joined at one end. The rod and the rails make a closed loop, and the field is at right angles to the loop. When the rod moves a distance $dx$, the area of the loop grows by $l\,dx$. So the flux changes by $B\,l\,dx$, and

The two methods agree, as they must. The flux method also shows the cause: the loop is getting larger.

The direction of the emf

Use Lenz's law. In the picture the field points into the page and the rod moves to the right. The loop grows, so the flux into the page grows. The induced field must point out of the page, and the current turns anticlockwise. So the current goes up through the rod.

You can check this with the force on a positive charge. Point the fingers of your right hand along the velocity, to the right. Curl them towards the field, into the page. The thumb points up. The force on a positive charge is up the rod, so the rod pushes current up.

Current, force and power

Let the rod and the rails have a total resistance $R$. The current is

A current in a rod in a field feels a force, $F = BIl$. By Lenz's law this force must oppose the motion. In the picture it points to the left. To keep the rod at a steady speed, you must pull it with an equal force to the right:

The power that you supply is $P = Fv$. The heat produced in the resistor each second is $I^2R$. Put in the numbers symbolically:

The two are equal. All the work that you do becomes heat. This is conservation of energy, and it shows again why Lenz's law could not be the other way round.

An example

A rod of length $0.50$ m slides at $2.0$ m/s on rails in a field of $0.40$ T. The whole circuit has a resistance of $1.0\ \Omega$. Find the emf, the current, the force that you apply and the power.

1. The emf. $\varepsilon = Blv = 0.40\ \text{T} \times 0.50\ \text{m} \times 2.0\ \text{m/s} = 0.40\ \text{V}$.
2. The current. We want $I$, so $I = \dfrac{\varepsilon}{R} = \dfrac{0.40\ \text{V}}{1.0\ \Omega} = 0.40\ \text{A}$.
3. The force. $F = BIl = 0.40\ \text{T} \times 0.40\ \text{A} \times 0.50\ \text{m} = 0.080\ \text{N}$.
4. The power. $P = Fv = 0.080\ \text{N} \times 2.0\ \text{m/s} = 0.16\ \text{W}$. The heat is $I^2R = (0.40\ \text{A})^2 \times 1.0\ \Omega = 0.16\ \text{W}$. The two agree.

Double the speed. The emf and the current double, and so does the force. The power grows four times, as the curve shows. Now double the field. The emf doubles, but the force and the power grow four times, because the field acts twice: once to make the current and once to push on it.

When the field is tilted

Only the part of the field at right angles to both the rod and its velocity makes an emf. Suppose a rod moves along horizontal rails in a field that makes an angle $\alpha$ with the horizontal. The part of the field along the rails does nothing. The vertical part, $B\sin\alpha$, does:

Two limits show this. If the field points along the rails, then $\alpha = 0$ and the emf is zero. If the field is vertical, then $\alpha = 90^\circ$ and $\varepsilon = Blv$.

The same idea applies to an aircraft wing. The field of the Earth has a vertical part and a horizontal part. A wing moves horizontally, so it cuts the vertical part $B_v$. An emf $B_v\,lv$ appears between the wing tips, with $l$ the span of the wing. A voltmeter in the aircraft cannot show it, because the wires of the voltmeter move through the same field.

A spinning rod

A metal rod of length $l$ turns about one end with angular speed $\omega$, in a field at right angles to its plane. The speed of a point at a distance $r$ from the pivot is $\omega r$. It grows from zero at the pivot to $\omega l$ at the free end. The average speed is $\omega l/2$, so the emf between the ends is

Check this with Faraday's law. In a short time $dt$ the rod sweeps a sector of area $\tfrac{1}{2}l^2\,\omega\,dt$. The flux that it cuts is $\tfrac{1}{2}Bl^2\omega\,dt$, and the rate is the same $\tfrac{1}{2}B\omega l^2$.

Practise with new numbers

Try it in Python

The program finds the emf, the current, the force and the power for a rod that moves at four speeds. It also finds the heat in the resistor, to compare with the power that you supply.

Common mistakes

• Using $Blv$ when the rod moves along the field or along its own length. Then no charge is pushed along the rod, and the emf is zero.
• Using the whole field when it is tilted. Use only the part at right angles to the rod and to its velocity.
• Forgetting that a current needs a closed circuit. A rod with no loop has an emf but no current and no force.
• Mixing up the two powers. The power that you supply, $Fv$, equals the heat in the resistor, $I^2R$. It is not the product of the emf and the force.
• Using the length in cm. The length must be in metres.

Remember these

• Motional emf: $\varepsilon = Blv$, when $B$, $l$ and $v$ are at right angles to one another.
• The free charges feel the force $qvB$. The emf is the electric field that balances it, times the length.
• On rails: $I = \dfrac{Blv}{R}$, force $F = \dfrac{B^2l^2v}{R}$ (opposing the motion), power $P = Fv = \dfrac{B^2l^2v^2}{R} = I^2R$.
• Tilted field: use the part of $B$ at right angles to the rod and to its velocity.
• Rod turning about one end: $\varepsilon = \tfrac{1}{2}B\omega l^2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A rod $0.60$ m long moves at $4.0$ m/s at right angles to its length and to a field of $0.35$ T. What is the emf between its ends?

2. The same rod is moved along its own length, with the same speed, in the same field. Is there an emf between its ends? Why?

3. A rod on rails, joined through a resistor, is pulled at a steady speed. The speed is doubled. By what factor do the emf, the current, the force and the power change?

4. A rod slides on two rails that are not joined to anything. Is there an emf? Is there a force on the rod that you must overcome?

5. A metal rod $0.80$ m long turns about one end at $20$ rad/s, in a field of $0.50$ T at right angles to its plane. What is the emf between its ends?