1. (a) Write down the formula for the
force on a straight wire of length L placed at right angles to a
magnetic field of flux density B.
(b) A pivot arrangement using a wire of length 15 cm is
placed in a magnetic field. When a current of 4 A is passed through the wire it
is found that a mass of 2.0 g must be hung from the wire to return it to a
horizontal position. g = 9.81 m s-2
Calculate the magnetic flux density
of the magnet.
2. A horizontal wire 6 cm long and mass 1.5 g is placed at
right angles to a magnetic file of flux density 0.5 T. Calculate the current
that must be passed through the wire so that it is self-supporting
3. A wire of length 0.5 m carrying a current of 2 A is
placed at right angles to a magnetic field of 0.2 T. Calculate the force on the
4. What is the force between two
wires placed 0.2 m apart in a vacuum each carrying a current of 5 A.
Questions 5 and 6:
A magnetic field exerts a force of 0.25 N on
an 8.0 cm length of wire carrying a current of 3.0 A at right angles to the
5. Calculate the force the same field would exert on a wire
20 cm long carrying the same current.
6. Calculate the force the same field would exert on three
insulated wires, each 20 cm long and held together parallel to each other, each
carrying a current of 3.0 A in the same direction.
7. (a) Using the ideas of forces on currents in
magnetic fields and the diagram explain the operation of a moving coil
(b) Calculate the force on the voice coil of
a loudspeaker that has a diameter of 2 cm, a 100 turns and carries a current of
20 mA if the field strength of the tubular magnet is 0.1T.
Answers and worked solutions
1(a) F = BIL
B = [2.0 x 10-3
x 9.81]/ [4 x 0.15] = 0.0327 T (Remember to change grams to kg and
centimetres to metres)
2 Weight of mass required = BIL
Therefore: 1.5 x 10-3
x 9.81 = 0.5 x I
x 0.06 giving I = 0.49 A
3 F = BIL = 0.2 x 2 x 0.5 = 0.2 N
4 F = moI2L/2pr = 4p x10-7 x 25
x ½p x 0.2 = 2.5 x10-5 N
5 For 20 cm length
6. If each wire carries 3.0 A this is
the same as effective current of 9.0 A,
so force F = 3 ´ 0.625 =
7(a) The wire of voice coil is at right angles to the field.
When a varying current from an amplifier current flows in the coil it
experiences a force that moves the coil backwards and forwards along the soft
(b) F = nBIL = 100 x 0.1 x 20 x 10-3
x p x 2 x 10-2 = 0.013 N
(Note: cm to m and
mA to A)
Questions 1-4 and 7 are taken from
Resourceful Physics Questions 5 and 6 are from Advancing Physics chapter 15,
TAP 412- 4: Forces on currents in magnetic fields µ o = 4π ´ 10 -7
Using the ideas of forces on currents in magnetic fields and the diagram explain the operation of a moving coil loudspeaker