Force on a current-carrying conductor

The force on a current

A current is a stream of moving charges. We know the field pushes on each moving charge, so it must push on the wire that holds them. Let us find the total.

Take a straight piece of wire of length $L$ and cross section $A$, in a field $B$. The wire has $n$ free charge carriers in each cubic metre, each with charge $q$ and drift speed $v_d$. Suppose the field is at right angles to the wire.

1. The number of carriers. The piece has a volume $AL$, so it holds $nAL$ carriers.
2. The force on one. $F_1 = qv_dB$.
3. The force on all. $F = nAL \times qv_dB = (nAqv_d)\,LB$.

Recall the drift of charges in a wire: $I = nAqv_d$. So the force is

If the wire makes an angle $\theta$ with the field, only the part of the length across the field counts, which is $L\sin\theta$. So

In vector form, $\vec{F} = I\,\vec{L} \times \vec{B}$. The vector $\vec{L}$ has the length of the wire and points along the current. The force is at right angles to the wire and to the field.

The unit follows: $B = \dfrac{F}{IL}$, so $1\ \text{T} = 1\ \text{N}/(\text{A m})$. A wire of length $1$ m with a current of $1$ A, at right angles to a field of $1$ T, feels $1$ N.

The direction of the force

You may use the right-hand rule of the first lesson, with $\vec{L}$ in place of $\vec{v}$. Point the fingers along the current. Turn them towards the field. The thumb gives the force.

Many books use Fleming's left-hand rule instead. Stretch the thumb, the forefinger and the middle finger of the left hand at right angles to each other. Let the forefinger point along the field, the middle finger along the current. The thumb then points along the force. Both rules give the same direction.

If you reverse the current, the force reverses. If you reverse the field, it reverses too. If you reverse both, it does not change.

A worked example

A straight wire $0.35$ m long carries $12$ A in a uniform field of $0.25$ T. The wire makes an angle of $40^\circ$ with the field. What is the force?

1. The formula. $F = BIL\sin\theta$.
2. The numbers. $F = 0.25\ \text{T} \times 12\ \text{A} \times 0.35\ \text{m} \times \sin 40^\circ = 0.25 \times 12 \times 0.35 \times 0.643\ \text{N} = 0.67\ \text{N}$.
3. The direction. At right angles to the plane that holds the wire and the field.

Make the angle $0^\circ$. Then make it $90^\circ$. Which gives the larger force? Then double the length and the current. By what factor does the force change?

A bent wire

Suppose a wire is not straight. Cut it into tiny straight pieces $d\vec{L}$. Each has a force $I\,d\vec{L} \times \vec{B}$. If the field is the same everywhere, we can add the pieces:

The sum of all the small vectors $d\vec{L}$ is the straight vector $\vec{L}_{\text{eff}}$ from the start of the wire to its end. So in a uniform field, a bent wire feels the same force as a straight wire between its two ends. The shape in between does not matter.

Two things follow.

• A half circle. A wire in the shape of a half circle of radius $R$ has its ends $2R$ apart. The force is $F = BI(2R)$, when the field is at right angles to the plane of the half circle. It is not $BI\pi R$.
• A closed loop. The start and the end are the same point, so $\vec{L}_{\text{eff}} = 0$. A closed loop in a uniform field feels no net force. The forces on its parts may still turn it, as we will see when we study loops of current.

A worked example. A wire goes $0.12$ m east and then $0.16$ m north. It carries $6.0$ A. A uniform field of $0.30$ T points straight up. Draw the wire on a map, with east to the right and north up the page. The field then points out of the page. Find the force on the wire.

1. The straight wire. The ends are joined by the vector $0.12$ m east and $0.16$ m north, of length $\sqrt{0.12^2 + 0.16^2} = 0.20$ m.
2. The size. $F = BIL_{\text{eff}} = 0.30\ \text{T} \times 6.0\ \text{A} \times 0.20\ \text{m} = 0.36\ \text{N}$.
3. The direction. Take east as $\hat{i}$, north as $\hat{j}$ and up as $\hat{k}$. Then $\vec{L}_{\text{eff}} \times \vec{B} \propto 0.12(\hat{i} \times \hat{k}) + 0.16(\hat{j} \times \hat{k}) = -0.12\,\hat{j} + 0.16\,\hat{i}$. The force is at right angles to the line joining the ends, towards the east and a little south.

A wire held up by the field

A magnetic force on a wire can balance its weight. Take a horizontal wire of length $L$ and mass $m$, at right angles to a horizontal field. If the current runs the right way, the force is upward. The wire floats when

This is how a current balance works: a wire of known mass is held up by the field, and the current shows the field.

A rod on threads, and a rod on rails

Hang a horizontal rod on two light threads, in a field that points up. When a current flows along the rod, the force is sideways. The rod swings out until the threads make an angle $\varphi$ with the vertical. There, the weight $mg$, the magnetic force $BIL$ and the pull of the threads are in balance. The two forces give

Make the current larger. How does the angle change? Does it ever reach $90^\circ$?

Now put the rod on two rails, in a field that points up. The force pushes the rod along the rails. A worked example. A rod of mass $0.060$ kg lies across rails that are $0.50$ m apart. The field is $0.12$ T, straight up, and the current is $7.5$ A.

1. The force. $F = BIL = 0.12\ \text{T} \times 7.5\ \text{A} \times 0.50\ \text{m} = 0.45\ \text{N}$.
2. The acceleration, without friction. $a = \dfrac{F}{m} = \dfrac{0.45\ \text{N}}{0.060\ \text{kg}} = 7.5\ \text{m/s}^2$.
3. With friction. If the coefficient of friction is $0.20$, the friction is $0.20 \times 0.060\ \text{kg} \times 9.8\ \text{m/s}^2 = 0.118\ \text{N}$. The net force is $0.45 - 0.118 = 0.332\ \text{N}$, so $a = 5.5\ \text{m/s}^2$.

Practise with new numbers

Try it in Python

The program finds how far a hanging rod swings for seven values of the current.

Common mistakes

• Using the length of a curved wire. In a uniform field, only the straight distance between the ends counts.
• Counting wire that is outside the field. Use the length that lies in the field.
• Using the angle from the wrong line. In $F = BIL\sin\theta$, the angle is between the wire and the field. At $0^\circ$ the force is zero.
• Getting the direction of the current wrong. The current goes from the positive terminal of the cell, outside the cell. Use that for $\vec{L}$.
• Thinking a closed loop is pulled as a whole. In a uniform field its net force is zero, since the pieces cancel.

Remember these

• $F = BIL\sin\theta$, or $\vec{F} = I\,\vec{L} \times \vec{B}$. Unit: $1\ \text{T} = 1\ \text{N}/(\text{A m})$.
• It is the sum of the Lorentz forces on the moving charges: $I = nAqv_d$.
• Direction: the right-hand rule, or Fleming's left-hand rule (forefinger field, middle finger current, thumb force).
• In a uniform field a bent wire feels the force of the straight wire between its ends. A half circle gives $BI(2R)$. A closed loop gives zero.
• A wire floats when $BIL = mg$. A hung rod swings to $\tan\varphi = \dfrac{BIL}{mg}$.

Check yourself

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

1. A $0.50$ m wire carries $8.0$ A along a field of $0.60$ T. What is the force on it?

2. A horizontal wire carries a current to the west. The field points straight down. In which direction is the force on the wire?

3. Why is the net force on a closed loop of current zero in a uniform field?

4. A wire in the shape of a quarter circle of radius $0.20$ m carries $5.0$ A in a uniform field of $0.30$ T at right angles to its plane. Find the force.