Field patterns of a long straight wire, a flat coil and a solenoid, and the right-hand grip rule

Magnetic flux density B and its unit, the tesla

Magnetic flux φ= BA and flux linkage Nφ, including a coil at an angle to the field

The force on a current-carrying conductor, F = BIlsin θ, and Fleming's left-hand rule

Measuring B with a current balance, including the uncertainty

The force on a moving charge, F = Bqvsin θ, for positive and negative charges

Circular paths in a uniform field and r = p/BQ; selecting one speed with crossed fields

1. A magnetic field is produced by moving charge and exerts a sideways force on moving charge, F = BIlsin θ on a wire and F = Bqvsin θ on a single charge, with the direction given by Fleming's left-hand rule.

2. Field lines run from N to S and never cross; closer lines mean a stronger field. The right-hand grip rule gives the direction of the field around a current.

3. Magnetic flux density B is the force per unit current per unit length on a wire at right angles to the field: 1 T = 1 N A⁻¹ m⁻¹.

4. Flux φ= BA (area at right angles to B) is in webers; flux linkage Nφ is in Wb turns.

5. Fleming's left-hand rule: First finger = Field, seCond finger = Current (conventional), thuMb = Motion (force). For a negative charge, point the second finger against its velocity.

6. The magnetic force on a charge is always at right angles to its velocity, so it does no work: in a uniform field the charge moves in a circle at constant speed with r = p/BQ.

A magnetic field is a region in which a moving charge, a current-carrying conductor or a magnetic material experiences a force. Field lines show the direction a free north pole would be pushed: out of N, into S. They never cross, and closer lines mean a stronger field.

Every current produces a magnetic field. Right-hand grip rule: grip the wire with your right hand, thumb along the conventional current; your fingers curl the way the field lines go. × means into the page (an arrow's tail) and • means out of the page (an arrow's point).

Bending the wire into a loop makes every part of it send field the same way through the middle. Many loops in a solenoid give a strong, nearly uniform field inside and a bar-magnet pattern outside.

Typical values: a fridge magnet is about 5 mT near its surface, a strong laboratory magnet 0.1–1 T, and the Earth's field about 50 µT. B is a vector: its direction is the direction of the field lines, and the density of the lines represents its size.

Flux measures how much field passes through an area, like the number of field lines threading it. Flux linkage counts every turn of a coil: N turns each enclosing φ link Nφ.

If the coil is tilted, only the component of its area at right angles to the field counts. With θ the angle between the field and the normal to the coil, φ= BAcos θ: maximum when the coil faces the field (θ= 0°), zero when its plane is parallel to the field (θ= 90°).

WORKED EXAMPLE · QUESTION

A square coil of side 4.0 cm has 250 turns. It is placed with its plane at right angles to a uniform field of flux density 0.15 T. Calculate (a) the flux through the coil, (b) the flux linkage, and (c) the flux linkage after the coil is rotated so that its normal makes an angle of 60° with the field.

SOLUTION

(a) Area in m²: A = 0.040² = 1.6×10⁻³ m², so φ= BA = 0.15 ×1.6×10⁻³ = 2.4 × 10⁻⁴ Wb

(b) Nφ= 250 ×2.4×10⁻⁴ = 0.060 Wb turns

(c) Only the component of the area at right angles to the field counts: Nφ= NBAcos 60° = 0.060 ×0.50 = 0.030 Wb turns

Common slip: forgetting to convert 4.0 cm to 0.040 m before squaring, which makes the area 10⁴ times too big.

A wire carrying a current across a magnetic field feels a force (the motor effect), at right angles to both the wire and the field: largest at 90°, zero when the wire is parallel to the field.

Fleming's left-hand rule

Hold the thumb and first two fingers of your left hand at right angles to each other. The First finger points along the Field (N to S), the seCond finger along the Current (conventional, + to −), and the thuMb gives the Motion, that is the direction of the force.

If the wire is not at 90° to the field, use the component of the current at right angles to the field.

Why is there a force? The combined field

The wire's own circular field adds to the magnet's field on one side of the wire and opposes it on the other. The wire is pushed from the region of strong combined field towards the region of weak field, which is exactly the direction Fleming's rule gives.

WORKED EXAMPLE · QUESTION

A straight wire lies in the plane of the page and carries a current of 4.0 A. A 0.30 m length of it is in a uniform field of flux density 0.080 T, directed to the right across the page. The wire points up and to the right, at 35° to the field. Calculate the force on the wire and state its direction.

SOLUTION

Use θ = the angle between the wire and the field: F = BIlsin θ= 0.080 ×4.0 ×0.30 ×sin 35° = 0.055 N (0.0551 N)

Direction: field to the right, component of the current up the page, so by Fleming's rule the force is into the page.

At 90° the force would be the maximum, 0.096 N; parallel to the field it would be zero.

WORKED EXAMPLE · QUESTION

A horizontal copper rod of mass 18 g hangs from two light, flexible leads. A 0.12 m length of the rod is in a horizontal magnetic field of flux density 0.30 T at right angles to the rod. (a) Calculate the current that makes the tension in the leads zero. (b) The current is reduced to 3.0 A. Calculate the tension in each lead.

SOLUTION

(a) Tension is zero when the upward magnetic force equals the weight:

BIl = mg, so I = mg/Bl = (0.018 ×9.81) ÷(0.30 ×0.12) = 4.9 A (4.91 A)

(b) Magnetic force at 3.0 A: F = 0.30 ×3.0 ×0.12 = 0.108 N. The leads support the rest of the weight, shared equally: 2T = 0.1766 - 0.108 = 0.0686 N, so each lead has T = 0.034 N

Reverse either the current or the field and the magnetic force points down: the leads become tighter, not slacker.

In a motor, the forces on opposite sides of a coil form a couple that turns it; in a loudspeaker, an alternating current makes the force on the coil reverse, so the cone vibrates. Two parallel wires each sit in the other's field: currents in the same direction attract, in opposite directions repel.

Set-up. A magnet with its poles facing, so that the field between them is close to uniform, stands on an electronic balance. A stiff straight wire, clamped from above so that it cannot move, passes between the poles at right angles to the field.

Newton's third law. If the field pushes the wire up, the wire pushes the magnet down with an equal force, so the balance reading increases by Δm, and F = Δm g with Δm in kg.

Method. Tare the balance with no current and measure Δm for a range of currents.

Reverse the current, repeat and average the sizes of the two readings: this cancels changes that do not reverse with the current, such as convection from the warm wire or zero drift.

Analysis. F = BIl, so a graph of Δm against I is a straight line through the origin with gradient Bl/g. So B = gradient × g/l, where l is the length of wire inside the field, roughly the width of the pole faces, measured with a ruler or calipers.

WORKED EXAMPLE · QUESTION

Use the graph above to determine the flux density between the poles. The length of wire in the field is 5.0 ± 0.1 cm, and the steepest and least steep acceptable lines differ from the best line by ±3%. Determine the percentage uncertainty in B and give the result with its absolute uncertainty.

SOLUTION

Gradient from a large triangle: 1.30 g ÷ 4.0 A = 0.325 g A⁻¹ = 0.325 × 10⁻³ kg A⁻¹.

Convert mass to force: gradient × g = F/I = Bl, so Bl = 0.325×10⁻³ ×9.81 = 3.19×10⁻³ N A⁻¹.

B = 3.19×10⁻³ ÷0.050 = 0.064 T (0.0638 T)

B = gradient × g ÷l, so add the percentage uncertainties: % in l = 0.1 ÷ 5.0 × 100 = 2.0%, % in B = 3% + 2% = 5%, so B = 0.064 ± 0.003 T

The intercept is 0.003 g, effectively zero: evidence that the zeroing worked and that there is no systematic error in the balance.

A current is moving charge, so each charge carrier in a wire feels part of the force BIl. The same force acts on a charge moving through empty space, at right angles to both the velocity and the field, and zero for a charge at rest or moving parallel to the field.

Direction. Use Fleming's left-hand rule with the second finger along the conventional current: along the velocity for a positive charge, and opposite to the velocity for a negative charge such as an electron.

WORKED EXAMPLE · QUESTION

An electron moves to the right across the page at 3.0 × 10⁷ m s⁻¹ through a uniform field of 2.0 mT directed out of the page. (a) Calculate the size of the force on the electron and its acceleration. (b) State the direction of the force.

SOLUTION

(a) Velocity at 90° to the field, so sin θ= 1: F = Bqv = 2.0×10⁻³ ×1.60×10⁻¹⁹ ×3.0×10⁷ = 9.6 × 10⁻¹⁵ N. Then a = F/m = 9.6×10⁻¹⁵ ÷9.11×10⁻³¹ = 1.1 × 10¹⁶ m s⁻²

(b) An electron moving right is a conventional current to the left; with the field out of the page, Fleming's rule gives a force up the page, at right angles to the velocity.

The weight of the electron, about 9 × 10⁻³⁰ N, is smaller by a factor of more than 10¹⁴, so gravity is always ignored for beams of particles.

The magnetic force is always at right angles to the velocity, so it does no work and the speed stays constant. A constant-size force always perpendicular to the velocity gives uniform circular motion: the magnetic force provides the centripetal force.

Bigger momentum gives a bigger circle; a stronger field or a bigger charge gives a tighter circle. Opposite charges curve opposite ways.

WORKED EXAMPLE · QUESTION

A proton and an alpha particle enter a uniform field of flux density 0.25 T at right angles to it, both at 2.2 × 10⁶ m s⁻¹. The mass of an alpha particle is 6.64 × 10⁻²⁷ kg and its charge is +2e. (a) Calculate the radius of each path. (b) Both enter at the same point and leave the field after a semicircle. Calculate the distance between their exit points.

SOLUTION

(a) Proton: r = mv/BQ = (1.67×10⁻²⁷ ×2.2×10⁶) ÷(0.25 ×1.60×10⁻¹⁹) = 0.092 m (9.19 cm). Alpha: r = (6.64×10⁻²⁷ ×2.2×10⁶) ÷(0.25 ×3.20×10⁻¹⁹) = 0.18 m (0.1826 m)

(b) Each exits one diameter from the entry point: 2 ×0.1826 - 2 ×0.0919 = 0.18 m

Ratio check without numbers: r ∝m/Q at the same speed, and 4m/2e is twice m/e, so the alpha particle's radius is about twice the proton's (1.99 times with these masses).

WORKED EXAMPLE · QUESTION

Singly charged positive ions pass between two parallel plates 8.0 mm apart with a p.d. of 480 V across them. A magnetic field of 0.050 T, at right angles to both the electric field and the ions' velocity, acts in the same region. Only ions for which the electric and magnetic forces cancel travel straight through. These ions, of mass 1.15 × 10⁻²⁶ kg, then enter a region where there is only a magnetic field of 0.40 T. (a) Calculate the speed of the ions that pass straight through. (b) Calculate the radius of their path in the 0.40 T field.

SOLUTION

(a) Electric field between the plates (4.3): E = V/d = 480 ÷8.0×10⁻³ = 6.0×10⁴ V m⁻¹. The forces balance, EQ = BQv, so the charge cancels: v = E/B = 6.0×10⁴ ÷0.050 = 1.2 × 10⁶ m s⁻¹

(b) r = mv/BQ = (1.15×10⁻²⁶ ×1.2×10⁶) ÷(0.40 ×1.60×10⁻¹⁹) = 0.22 m (0.216 m)

Faster ions feel a bigger magnetic force and are deflected one way; slower ions are deflected the other way by the unchanged electric force. Only one speed gets through, whatever the charge or mass.

A velocity component along the field is unaffected, so the particle spirals (a helix) along the lines. A particle losing energy in a detector has decreasing momentum, so its track spirals inwards, showing which way it travelled.

2 Circular motion and 4.3 Electric fields

The magnetic force Bqv plays the role of the centripetal force mv²/r, which is how r = p/BQ is derived. Compare the two fields: an electric force EQ acts on any charge, along the field lines, and can do work; a magnetic force acts only on moving charge, sideways, and does no work. Balancing them selects a single speed.

6 Electromagnetic induction

Flux φ= BA and flux linkage Nφ, defined here, are exactly what must change to induce an e.m.f.

7 Atomic structure and particle accelerators

Cyclotrons bend particles round with a magnetic field, and detectors find momentum from the radius of curvature of a track using r = p/BQ.

1. A beam of protons travels parallel to the field lines of a uniform magnetic field. What force does the field exert on them, and why?

2. A coil of 500 turns and area 2.0 cm² is placed with its plane at right angles to a field of 0.30 T. Calculate the flux linkage.

3. A 0.25 m length of wire carrying 6.0 A lies at right angles to a field of 0.12 T. Calculate the force on it. What is the force if the wire is turned to lie along the field?

4. A proton moves up the page through a field directed into the page. In which direction is the force on it? What would change for an electron?

5. An electron moves at 5.0 × 10⁶ m s⁻¹ at right angles to a field of 1.0 mT. Calculate the radius of its path, and explain why its speed does not change.

6. Two long parallel wires carry currents in opposite directions. Explain whether they attract or repel.

1. None. F = Bqvsin θ with θ= 0, so sin θ= 0: there is no component of the velocity across the field.

2. Nφ= NBA = 500 ×0.30 ×2.0×10⁻⁴ = 0.030 Wb turns.

3. F = 0.12 ×6.0 ×0.25 = 0.18 N. Along the field, sin 0° = 0, so the force is zero.

4. Fleming's left-hand rule (field into the page, current up the page) gives a force towards the left of the page. An electron moving the same way is equivalent to a current down the page, so the force on it is towards the right.

5. r = mv/BQ = (9.11×10⁻³¹ ×5.0×10⁶) ÷(1.0×10⁻³ ×1.60×10⁻¹⁹) = 0.028 m (2.8 cm). The force is always perpendicular to the velocity, so it does no work and the kinetic energy is constant.

6. Each wire is in the field of the other. Applying the grip rule to one wire and then the left-hand rule to the other shows the force on each wire points away from the other, so they repel. (Currents in the same direction attract.)