The Motor Effect

Solution

1. 1

   Identify the variables: B = 0.50 T, I = 6.0 A, L = 0.15 m.

   $F = BIL$

2. 2

   Substitute.

   $F = 0.50 \times 6.0 \times 0.15$

3. 3

   Evaluate.

   $F = 0.45\,\text{N}$

Solution

1. 1

   Rearrange the equation for I.

   $I = \dfrac{F}{BL}$

2. 2

   Substitute.

   $I = \dfrac{0.24}{0.6 \times 0.08}$

3. 3

   Evaluate.

   $I = 5.0\,\text{A}$

Solution

1. 1

   Point your LEFT first finger north (field).

2. 2

   Point your second finger east (current).

3. 3

   Your thumb points downward — the wire experiences a downward force.

Solution

1. 1

   The motor-effect force depends on the component of B perpendicular to the wire.

2. 2

   If the wire is parallel to B, the perpendicular component is zero.

3. 3

   Therefore F = 0. Equivalently, F = BIL sin θ with θ = 0 gives F = 0.

Solution

1. 1

   Convert to SI units: L = 0.050 m; I = 0.250 A; B = 0.080 T.

2. 2

   Apply F = BIL.

   $F = 0.080 \times 0.250 \times 0.050$

3. 3

   Evaluate.

   $F = 1.0\times10^{-3}\,\text{N} = 1.0\,\text{mN}$

Solution

1. 1

   Identify the field direction: N is on the left, S on the right, so B points from left to right (N → S).

2. 2

   On the near side, current flows left to right. First finger (left hand) right (field), second finger right (current). Hmm — they're parallel. Re-read: the near side current flows toward the viewer? Re-orient: current flows along the wire, perpendicular to B. With first finger right and second finger toward the viewer, the thumb points downward.

3. 3

   The near (front) side of the coil is pushed downward; the far side is pushed up. The coil rotates so the front goes down and the back rises.

Solution

1. 1

   Without a commutator, the current in the coil would always flow the same way relative to the battery.

2. 2

   After half a turn, the side of the coil that was on the left would now be on the right. The force on it (still up or down) would now act in the WRONG direction to keep the coil spinning the same way; the forces would slow the coil and reverse it.

3. 3

   The split-ring commutator swaps the connections every half-turn, reversing the current in the coil so the force on each side keeps pushing it round in the same direction. The coil rotates continuously.

Solution

1. 1

   Increase the current in the coil (higher voltage supply or lower resistance).

2. 2

   Increase the strength of the magnetic field (stronger permanent magnets or a stronger electromagnet).

3. 3

   Increase the number of turns on the coil.

4. 4

   Wrap the coil around a soft iron core to concentrate the magnetic field through it.

Solution

1. 1

   Method 1: reverse the current by swapping the battery terminals — both forces flip and the coil spins the other way.

2. 2

   Method 2: reverse the magnetic field by swapping the N and S poles — both forces flip again.

3. 3

   If both are reversed at the same time, the two flips cancel: the force on each wire returns to its original direction and the motor spins the same way it did before.

Solution

1. 1

   Useful transfer: electrical energy from the supply → kinetic energy of the rotating coil and drill bit.

2. 2

   Wasted transfers: heat (mostly from resistance heating of the coil and friction in the brushes/bearings) and a small amount of sound.

3. 3

   The motor warms up because the current through the coil dissipates power as heat via resistance (P = I²R) and because of friction between the brushes and commutator.

Solution

1. 1

   The voice coil sits in the radial magnetic field of a permanent magnet, so each turn of the coil carries current at right angles to the field.

2. 2

   By the motor effect (F = BIL), the coil experiences a force along its axis. As the AC reverses direction, the force reverses too.

3. 3

   The coil therefore oscillates back and forth at the frequency of the AC. It is glued to a paper cone, which vibrates with it.

4. 4

   The vibrating cone pushes and pulls air molecules, creating alternating compressions and rarefactions — a longitudinal sound wave of the same frequency as the AC.

Solution

1. 1

   (a) The frequency of the AC sets the frequency at which the coil oscillates and the cone vibrates. The frequency of the sound wave equals the AC frequency, so higher AC frequency → higher-pitched sound.

2. 2

   (b) The amplitude of the AC sets the size of the current I. By F = BIL, larger I means a larger force on the coil, so the coil and cone move further. Larger cone displacement produces larger pressure swings → louder sound.

Solution

1. 1

   DC produces a constant current in one direction, so the motor-effect force F = BIL acts in one direction only.

2. 2

   The cone is pushed outward (or pulled inward, depending on the polarity) and stays there — it does not vibrate.

3. 3

   Without vibration there are no compressions and rarefactions in the air, so no sound wave is produced. The student hears only a faint 'click' as the cone moves when the circuit closes (or breaks).

Solution

1. 1

   Headphone diaphragms are tiny and light, so only a small force (and therefore a small current) is needed to move them.

2. 2

   Sound waves spread out from a source; in headphones the diaphragm is millimetres from the eardrum, so almost all the sound energy reaches the ear.

3. 3

   A loudspeaker has to fill a whole room — most of the sound energy spreads sideways or is absorbed by walls, so much more power is needed at the source to achieve a comparable loudness at the listener's ear.

Solution

1. 1

   Force on each turn: F = BIL with B = 0.40 T, I = 0.50 A, L = 0.060 m.

   $F_{\text{turn}} = 0.40 \times 0.50 \times 0.060$

2. 2

   Evaluate per turn.

   $F_{\text{turn}} = 0.012\,\text{N}$

3. 3

   Multiply by N = 50 turns for the total force on the coil.

   $F = 50 \times 0.012 = 0.60\,\text{N}$