HSC Physics Module 6 Electromagnetism Questions with Solutions

• Motion of charges in electric fields

• Magnetic field produced by current carrying wires

• Magnetic force on a moving charge
• Magnetic force on a current carrying conductor
• Magnetic force between parallel conductors

• Motor effect in DC motors
• The roles of the main components in a DC motor
• Torque produced by a DC motor
• Improving torque characteristics

• Magnetic flux
• Faraday’s Law of Induction
• Lenz’s Law

• Eddy current
• Electromagnetic braking
• AC induction motor

• Back EMF in a DC motor

• Main components in AC and DC generators
• EMF in AC and DC generators
• Maximising EMF in AC or DC generators

• Power loss in electrical transmission
• Transformers

• Replace the single coil of the motor with several turns
• Replace the permanent plane magnets with radial magnets

Account for the difference in time taken for the magnet to fall.

• The type of a transformer
• Ratio of the number of turns in the primary and secondary coils

Given: q₁ = q₂ = q and v₁ = v₂ = v

Since r=\frac{mv}{qB} =(\frac{v}{qB}) \times v , r is directly proportional to m.

Therefore a larger radius means the particle possesses a larger mass.

F_{QR}=F_{PQ}=F \text{repulsive } since the current passing through the conductors are the same and the distance of separation between conductors Q and R is the same.

F_{PR}= \frac{F}{2} \text{ attractive } since the distance between conductors P and R is double the distance between conductors Q and R.

Therefore, the forces acting on conductor P are:

• F to the left
• \frac{F}{2} to the right

Hence the net force acting on conductor P is \frac{F}{2} to the left

In the formula for torque \tau=nIAB sin \theta represents the angle between the normal of the coil and the direction of the magnetic field. In this case it is 60̊ (or 120); it is 90 degrees away from the angle of the coil and magnetic field.

On the other hand,  \phi in the magnetic force formula  F=ILB sin \phi represents the angle the wire makes with the magnetic field – it is perpendicular to the magnetic field at all times.

The force acting on the coil AB and CD are always maximum since the current flowing through the coils is at right angles to the magnetic field. The direction of the force is reversed once every half turn by the commutator.

The torque produced by the coil is given by \tau=nIAB sin \theta . Since radial magnets are used, the angle between the magnetic field and the perpendicular area vector is at right angles in most positions. Hence the torque is at maximum \tau=nIAB at almost all orientations.

In a DC motor the commutator, 

• Reverses the direction of the current into the rotor coils once every half rotation
  
• Ensures the direction of torque remains in the same direction for continuous rotation

Part (b)

• \tau=nIAB sin \theta = 100 \times 4 \times (0.2 \times 0.1) \times 2.0 \times 10^{-2} sin 90 = 0.16 \ Nm
• The direction of the torque produced is clockwise.

Part (c)

• The mass should be attached on the left hand side, coil AB.
• \tau_coil = \tau_mass
• 8 = mg \times d
• m= \frac {0.16}{gd} = \frac{0.16}{9.8 \times 0.05}= 0.327 \ kg

Part (d)

• The torque increases as the number of turns of the coil increases.
• The torque becomes steady as due to radial magnetic field produced

• When viewed from the right, the current flows through the solenoid in an anti-clockwise direction.
• This will produce an increasing magnetic field to the right to oppose the increasing magnetic field to the left by the incoming magnet.

Part (b)

When the eddy current flows in an anticlockwise direction as shown, it produces an opposing magnetic field out of the page to cancel the increasing magnetic flux the sheet experiences as it enters the magnetic field. Consequently, it will establish opposing magnetic polarities which oppose the falling motion.

• Faraday’s Law states that an EMF is induced in a circuit when it experiences a time rate of change of magnetic flux.
• Furthermore, Lenz’s Law states that the direction of induced EMF is such that the current produced creates a magnetic flux that opposes the change in magnetic flux.
• The copper tube is conducting and will experience an EMF as the magnet falls through. Eddy currents are produced around the tube so that the magnetic flux created by them opposes the change in flux due to the falling magnet. As a result, the bottom end of the magnet is repelled upward and the top end is attracted upward, effectively slowing the magnet’s motion.
• The plastic tube is not conducting so it does not experience an EMF. No current is induced and thus there is no opposing magnetic flux to slow the magnet’s motion. As such, the magnet falls more quickly through the plastic tube

Refer to the flowchart for the explanation.

• Slip rings.
• They make continuous and sliding contact with the coil without reversing the direction of the induced current in the coil.

Part (b)

• The brightness of the light bulb flashes twice in the first revolution and four times in the second revolution with greater brightness.

Part (c)

Part (d)

• Modification: Introduce a set of perpendicular coils. OR Introduce a set of perpendicular magnets / radial magnets.
• Explanation:  A set of coils perpendicular to the existing coil’s orientation ensures that either coil is always experiencing a torque. Therefore, the overall torque and thus current produced by the generator will never be zero and will fluctuate less.

• Output voltage is AC.
• The direction of the induced currents switch when the polarity of the magnet reverses, therefore the output voltage is not unidirectional.

Part (b)

• Step up transformer
• \frac{N_p}{N_s}=\frac{V_p}{V_s} = \frac{23}{330}=0.070

Part (b)

• P=I_sV_s
• 700 \times 10^6 = I_s \times 300 \times 10^3
• I_s = 2.12 \times 10^3 \ A

Part (c)