9.3 Magnetic fields — Practise

Work through these on your own. Decide whether each question needs the force on a wire or the force on a moving charge, and use Fleming's left-hand rule to check every direction. Write the unit of every answer. In the questions that ask for a reason, write your answer in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A straight wire of length 0.60 m carries a current of 2.5 A at right angles to a magnetic field of flux density 0.35 T. Find the force on the wire.
The wire in question 1 is turned so that it makes an angle of 55 degrees with the field. Find the force on it now.
An electron moves at 3.0e6 m/s at right angles to a magnetic field of flux density 0.15 T. Find the magnetic force on it.
Find the radius of the circular path of the electron in question 3. The mass of an electron is 9.11e-31 kg.
Find the time the electron in question 3 takes to complete one full circle.
A coil of area 2.0e-3 m^2 is placed at right angles to a magnetic field of flux density 0.45 T. Find the magnetic flux through one turn.
The coil in question 6 has 250 turns. Find the flux linkage.
A wire of length 1.0 m carrying a current of 1.0 A at right angles to a field feels a force of 1.0 N. Show that the flux density is 1.0 T.
Explain why a charged particle moving along the lines of a magnetic field feels no magnetic force.
The radius of a charged particle in a magnetic field is 0.070 m. The flux density is measured with an uncertainty of 1.5% and the speed with an uncertainty of 0.5%. Find the absolute uncertainty in the radius.
A calculation gives a magnetic force of 9.6154e-14 N. Write the answer to two significant figures.
A magnetic field has a flux density of 3.0 mT. Write this in tesla.
A xenon ion of mass 2.18e-25 kg and charge +1.60e-19 C is accelerated from rest through 1.20 kV. It then enters a magnetic field of flux density 0.25 T at right angles to its velocity. Find the radius of its path.
State which equation you would use to find the radius of the path of an ion in a mass spectrometer, and why.

Answers and concise solutions

1. The force on the wire

Answer: $F = BIL = 0.35 \times 2.5 \times 0.60 = 0.53\,\text{N}$ .

2. The force on the wire

Answer: $F = BIL\sin\theta = 0.35 \times 2.5 \times 0.60 \times \sin 55^{\circ} = 0.43\,\text{N}$ .

3. The magnetic force

Answer: $F = BQv = 0.15 \times 1.60 \times 10^{-19} \times 3.0 \times 10^{6} = 7.2 \times 10^{-14}\,\text{N}$ .

4. The radius

Answer: $r = \dfrac{mv}{BQ} = \dfrac{9.11 \times 10^{-31} \times 3.0 \times 10^{6}}{0.15 \times 1.60 \times 10^{-19}} = 1.1 \times 10^{-4}\,\text{m}$ .

5. The period

Answer: $T = \dfrac{2\pi m}{BQ} = \dfrac{2\pi \times 9.11 \times 10^{-31}}{0.15 \times 1.60 \times 10^{-19}} = 2.4 \times 10^{-10}\,\text{s}$ .

6. The magnetic flux

Answer: $\Phi = BA = 0.45 \times 2.0 \times 10^{-3} = 9.0 \times 10^{-4}\,\text{Wb}$ .

7. The flux linkage

Answer: $N\Phi = NBA = 250 \times 0.45 \times 2.0 \times 10^{-3} = 0.23\,\text{Wb}$ .

8. The flux density

Answer: Rearranging $F = BIL$ gives $B = \dfrac{F}{IL} = \dfrac{1.0}{1.0 \times 1.0} = 1.0\,\text{T}$ . This is the definition of the tesla.

9. Explain why a charged particle moving along the lines of a magnetic field feels no magnetic force.

The angle $\theta$ between the velocity and the field is zero.
The force depends on $\sin\theta$ , and $\sin 0^{\circ} = 0$ .
Only the component of the velocity at right angles to the field gives a force.
A charge moving along the field has no component of velocity at right angles to it, so the force is zero.

10. The absolute uncertainty

Answer: The radius is $r = \dfrac{mv}{BQ}$ . A quotient adds percentage uncertainties, so the percentage uncertainty is $1.5 + 0.5 = 2.0\%$ . The absolute uncertainty is $\dfrac{2.0}{100} \times 0.070 = 1.4 \times 10^{-3}\,\text{m}$ . So $r = 0.070 \pm 0.001\,\text{m}$ .

11. The force to two significant figures

Answer: $9.6154 \times 10^{-14}\,\text{N}$ is $9.6 \times 10^{-14}\,\text{N}$ to two significant figures.

12. The flux density in tesla

Answer: $3.0\,\text{mT} = 3.0 \times 10^{-3}\,\text{T}$ , or 0.0030 T.

13. The radius

Answer: The kinetic energy is $E_{k} = eV = 1.92 \times 10^{-16}\,\text{J}$ , so $v = \sqrt{\dfrac{2E_{k}}{m}} = 4.20 \times 10^{4}\,\text{m}\,\text{s}^{-1}$ . Then $r = \dfrac{mv}{BQ} = \dfrac{2.18 \times 10^{-25} \times 4.20 \times 10^{4}}{0.25 \times 1.60 \times 10^{-19}} = 0.23\,\text{m}$ .

14. State which equation you would use to find the radius of the path of an ion in a mass spectrometer, and why.

The magnetic force on the moving ion is $F = BQv$ .
The force needed for circular motion is $F = \dfrac{mv^{2}}{r}$ .
Set the two expressions equal, because the magnetic force provides the centripetal force.
Rearranging gives $r = \dfrac{mv}{BQ}$ .