9.3 Magnetic fields — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Draw the field, the current and the force for every question that asks for a direction, and use Fleming's left-hand rule.

Warm-up: remember

A wire carries a current of 4.0 A. Find the charge that passes a point in the wire in 30 s.
A particle moves in a circle of radius 0.50 m at a constant speed of $3.0\,\text{m}\,\text{s}^{-1}$ . Find its centripetal acceleration.
Resolve a force of 20 N acting at $30^{\circ}$ to the horizontal into a horizontal and a vertical component.

Answers

$Q = It = 4.0 \times 30 = 120\,\text{C}$ .
$a = \dfrac{v^{2}}{r} = \dfrac{3.0^{2}}{0.50} = 18\,\text{m}\,\text{s}^{-2}$ .
Horizontal $= 20\cos 30^{\circ} = 17\,\text{N}$ ; vertical $= 20\sin 30^{\circ} = 10\,\text{N}$ .

If question 1 goes wrong, revisit charge and current in Section 5.1. If question 2 goes wrong, revisit circular motion in Section 3.5. If question 3 goes wrong, revisit resolving vectors in Section 2.2.

Practise with help

1. A straight wire of length 0.50 m carries a current of 2.0 A at right angles to a magnetic field of flux density 0.40 T. Find the force on the wire.

Hint

The wire is at right angles to the field, so no angle factor is needed.

Next step

Use $F = BIL$ with the flux density in tesla, the current in amperes and the length in metres.

Full working

F = BIL = 0.40 \times 2.0 \times 0.50 = 0.40\,\text{N}

The force is $0.40\,\text{N}$ , at right angles to both the wire and the field.

2. The wire in question 1 is turned so that it makes an angle of $40^{\circ}$ with the field. Find the force on it now.

Hint

Only the component of the wire at right angles to the field carries the force.

Next step

Use $F = BIL\sin\theta$ with $\theta = 40^{\circ}$ , and check that the answer is smaller than the answer to question 1.

Full working

F = BIL\sin 40^{\circ} = 0.40 \times 0.643 = 0.26\,\text{N}

The force is $0.26\,\text{N}$ . It is smaller than when the wire was at right angles, because only part of the wire is at right angles to the field.

3. A proton moves at $2.0 \times 10^{6}\,\text{m}\,\text{s}^{-1}$ at right angles to a magnetic field of flux density 0.30 T. Find the force on the proton.

Hint

A single moving charge obeys a different equation from a current-carrying wire.

Next step

Use $F = BQv$ with the charge on a proton, $Q = 1.60 \times 10^{-19}\,\text{C}$ .

Full working

\begin{aligned} F &= BQv = 0.30 \times 1.60 \times 10^{-19} \times 2.0 \times 10^{6} \\ &= 9.6 \times 10^{-14}\,\text{N} \end{aligned}

The force is $9.6 \times 10^{-14}\,\text{N}$ , at right angles to the velocity and to the field.

4. Find the radius of the circular path of the proton in question 3. The mass of a proton is $1.67 \times 10^{-27}\,\text{kg}$ .

Hint

The magnetic force provides the centripetal force.

Next step

Set $BQv = \dfrac{mv^{2}}{r}$ and rearrange for $r$ , or use $r = \dfrac{mv}{BQ}$ directly.

Full working

\begin{aligned} r &= \frac{mv}{BQ} = \frac{1.67 \times 10^{-27} \times 2.0 \times 10^{6}}{0.30 \times 1.60 \times 10^{-19}} \\ &= \frac{3.34 \times 10^{-21}}{4.80 \times 10^{-20}} = 7.0 \times 10^{-2}\,\text{m} \end{aligned}

The proton travels in a circle of radius 7.0 cm.

5. A coil of 150 turns has an area of $3.0 \times 10^{-3}\,\text{m}^{2}$ . It is placed at right angles to a magnetic field of flux density 0.20 T. Find the magnetic flux through one turn and the flux linkage.

Hint

Flux is for one turn; flux linkage includes all the turns.

Next step

Find $\Phi = BA$ first, then multiply by $N = 150$ .

Full working

\Phi = BA = 0.20 \times 3.0 \times 10^{-3} = 6.0 \times 10^{-4}\,\text{Wb}

N\Phi = 150 \times 6.0 \times 10^{-4} = 0.090\,\text{Wb}

The flux through one turn is $6.0 \times 10^{-4}\,\text{Wb}$ and the flux linkage is $0.090\,\text{Wb}$ .

6. Explain why the speed of a charged particle moving in a magnetic field stays constant, even though a force acts on it.

Hint

Think about the angle between the force and the velocity.

Next step

Say what the force is at right angles to, and what work done depends on.

Full working

The magnetic force is always at right angles to the velocity of the particle. Work done is force multiplied by the distance moved in the direction of the force. The particle does not move in the direction of the force, so the force does no work on it. Its kinetic energy does not change, so its speed stays constant. The force only changes the direction of the velocity.

7. Explain how a mass spectrometer uses a magnetic field to separate the isotopes of an element.

Hint

Isotopes have the same charge but different masses.

Next step

Start from $r = \dfrac{mv}{BQ}$ and say which quantities are the same for each isotope and which one differs.

Full working

The ions are all given the same charge and accelerated to the same speed. They enter the magnetic field at right angles. The radius of the circular path is $r = \dfrac{mv}{BQ}$ . The charge $Q$ , the speed $v$ and the flux density $B$ are the same for every isotope, so the radius depends on the mass $m$ . A heavier isotope travels in a larger circle and lands further along the detector, so the isotopes are separated.

8. A wire carrying a current is placed at an angle of $35^{\circ}$ to a magnetic field. Explain how you would find the force on the wire.

Hint

Only one component of the current feels a force.

Next step

Say which equation you would use, and then how you would find the direction.

Full working

Find the component of the wire that is at right angles to the field, or use $F = BIL\sin\theta$ with $\theta = 35^{\circ}$ . Then use Fleming's left-hand rule: point the first finger along the field, the second finger along the current, and the thumb gives the direction of the force. The force is at right angles to both the wire and the field.