# sa6 revision solutions ```YEAR 12 TERM 4 ELECTROMAGNETISM PHYSICS REVISION.
Q1. Calculate the force on a 5 cm long wire carrying 3 mA of current in a magnetic field of 10 mT if (a) the
wire is perpendicular to the field and (b) the wire is at an angle of 30o .
(a.) F = BIL = 1.5 &times; 10-6N (b.) F = BILsinθ = 7.5 &times; 10-7N
Q2. A vertical wire is in a north pointing magnetic field. If the wire is forced toward the west , in which
direction is the current flowing in the wire?
B
F
wire, current: positive z directoin
Q3. Calculate the magnetic field strength 3 cm from a wire carrying 2A of current.
𝑩=
𝒌𝑰
𝒓
; B = 1.33 &times; 10-5T
Q4. Calculate the field inside a solenoid 50 cm long (60 loops) , carrying 10 mA of current.
B = 2πknI; B = 1.51 &times; 10-6T
Q5. Write a paragraph on the operation of a galvanometer.
A galvanometer operates like a motor. A current passes through a coil wrapped in square loop
(with a rotating axis attached to a needle), so that two sections of the loop are at right angles to the field
direction supplied by a magnet either side of the loop. The right hand rule for a current carrying wire
inside a magnetic field can then be to described the direction in which the loop will spin. See diagram
below….
B
N
S
Q6. Calculate the period of a pendulum of length 2m. ???????????????????? IGNORE
Q7. Find the magnetic flux through a loop of radius 5 cm sitting in the earth’s magnetic field (B=5x10-5). The
loop and the field are perpendicular.
φ = BA = 3.93 &times; 10-7 Wb
Q8. Lenz’s Law states ......................................
A change in flux through a loop of wire induces an emf in the loop. The direction of the current produced
by the induced emf is such that the flux generated by the current tends to oppose the original change in
flux through the loop.
Q9. A 3m long rod is dropped in a field of 30mT (north). The rod reaches a speed of 5 m/s. Calculate the
induced emf on the rod. Assume the rod and field are at right angles.
motional induced emf = BLv = 0.45V
Q10. A transformer has a ratio of 10:1 for the number of turns in the primary coil to the secondary coil. If the
emf in the primary coil is 1000V , calculate the emf in the secondary coil. Is the transformer a step-up or stepdown transformer.
100V, step-down
Q11. State Lenz’s Law. Use it to describe the direction of the current in the ring in the following diagram :
Using the direction of the coil wire and the right hand rule for solenoids, a south pole will be produced at
the top of the coil. Because of Lenz’s Law, this will generate a south pole beneath the ring. Using the
solenoid right hand rule again, with thumb pointing upwards, the fingers will show the current direction
as anticlockwise if looking down on top of it.
Q12. Draw a horseshoe magnet and a bar magnet showing the magnetic field lines. Look in text.
Q13. (a.) Work out the size of the magnetic flux density 4cm from this wire and also show using the symbols
 and  the direction of the field.
 I = 15mA
&times;
&times;
&times;
&times;
𝑩=
𝒌𝑰
𝒓
,
B = 7.5 &times; 10-8T
•
•
•
•
(b.) (i.) Work out the poles of the two magnets shown if the current carrying wire is being pushed out of the
page.
(ii.) Also work out the size of the field if the wire is pushed out with a force of 12N at 600 to the field.
The wire is 30cm long.
I = 2A
S
F = BILsinθ, 𝑩 =
𝑭
60o
N
; B= 23.1T
𝑰𝑳 𝐬𝐢𝐧 𝜽
Q14. A ring of diameter 18mm is placed on a horizontal surface where the Earth’s magnetic field penetrates
downwards at an angle of 200 to the vertical. BE = 4.5  10-5 T. What is the magnetic flux through the
loop?
200
φ = BAcosθ, φ = 1.076 &times; 10-8Wb
Q15. (refer to the following diagram)
(a.) A solenoid which is 20cm long, contains 200 turns. If a current of 300mA passes through it , what
will be the size of the magnetic field in the solenoid?
B = 2πnkI; B = 3.77 &times; 10-4T
(b.) If the ring from question 14 is placed horizontally inside the solenoid, what will be the magnetic flux
inside the ring? 0Wb as there is no perpendicular component of B to the area of the ring.
What will it be if the ring is placed in vertically? φ = BA = 9.6 &times; 10-8Wb.
What will be the induced EMF if the ring is then rotated through 1800 every two seconds? This is difficult
to visualise. Consider the front of the ring. It rotates through 1800, this means that front of the ring is
now at the back. The relative direction of the magnetic field lines have now been reversed. Being
vectors, if the direction has reversed we can say one of them is in a negative direction (arbitrarily
chosen).
Before rotation, B field lines pointing into right hand side
The ring turns around so what was the right hand side is now on the left, even though the field lines have
not changed direction, relative to the side of the ring we are considering, they are now pointing out of
the ring instead of into. Looking at it another way, this would have the same effect as keeping the ring
still but reversing the direction of the magnetic field every 2 seconds.
∆∅
(𝝓 − 𝝓𝒊 )
∴ induced Emf = 𝑵 𝚫𝒕 = 1 &times; 𝒇 𝟐
= (BAf – BAi)/2 = (9.6 &times; 10-8 – 9.6 &times; 10-8)/2 = 9.6 &times; 10-8V
(c.) If the ring is fixed vertically and the current is switched to A.C. with a frequency of 50Hz, what will be
the induced EMF inside the ring?
This means the current through the solenoid is AC, making its magnetic field alternate at 50 Hz.
∴ time for one full change of B direction is 0.02s. Using the explanation form part (b.) above, the
direction of the field lines is changing(therefore changing the flux which induces a voltage(emf))
𝚫𝝓
𝒊𝒏𝒅𝒖𝒄𝒆𝒅 𝑬𝒎𝒇 = 𝑵 𝚫𝒕 , = (1 &times; 2 &times; 9.6 &times; 10-8)/0.02 = 9.6 &times; 10-6V
Q16. A transformer changes 110V to 12V. What is the ratio of the secondary turns to the primary turns?
What sort of current must be used for a transformer and why?
The order of the ratio is important, you are asked for secondary:primary, so NS:NP = VS:VP = 12:110 =
6:55. A.C. must be used so that the magnetic field in the primary coil which induces emf in the secondary
coil, will be constantly changing, as induced emf can only be produced when there is a change in
magnetic flux.
Q17. If the plan view of a coil in a galvanometer shown below, travels in an anti-clockwise direction , show
the direction of the current.
N
S
Using the right hand rule for a current carrying wire in a magnetic field, the left hand side of the coil is
being forced downward and the right hand side is being pushed upwards. You can refer to the diagram
in the text on galvanometers to visualise this. This explanation can also be used to explain the operaton of
a galvanometer in question 5.
So, considering the left hand side of the coil, Force(palm) is downward, Magnetic field (fingers) is to the
right(N to S) so therefore thumb(current) is pointing into the page (shown by a cross or arrows drawn on
the wires running from top right to lower left on top of the coil diagram.
Just to check, consider the right hand side of the coil rotating upwards (force- palm pushing up)
fingers(field) to the right as before &amp; therefore thumb(current) pointing upwards – a dot- coming out of
the page or positive z direction.
Q18. The Earth’s magnetic field travels in a direct southerly direction and goes into the ground at an angle of
30o to the horizontal in a certain place. What would be the induced EMF in a 2m car axle if the car is
travelling due West at 60km/h? Which wheel would be the positive electrode? (BE = 4.5  10-5T)
B
S
Bsin300
300
ground
Bcos300
Using the right hand rule for motional induced emf (thumb for velocity of moving conductor, palm for
direction of emf (+ to -), fingers for magnetic field direction)
Bcos300
Top view
v
axle
So, considering the side view, and
using the right hand rule, the emf
direction is out of the page, which
considering the Top view, is the
southerly direction.
v
Side view:
As the axle (and therefore, the emf
direction) is parallel to the field line,
we need to consider the vertical
component of the field vector, so a side
view must be considered. The B vector
here, represents the horizontal
component of the field, parallel with
the Earth’s surface.
Bsin300
Induced Emf = BLvsin300 = 7.5 &times; 10-4V
wheel
Q19. A beta particle enters a vertical magnetic field at right angles travelling at 2  107 ms-1 .
 




























If the field is equal to 4  10-1 T work out the radius of curvature of the particle’s path and show the
direction of the particle’s path.
A beta particle is an electron (negative charge) so the thumb must be going upwards.
𝒎𝒗𝟐
𝒓
= 𝑩𝒒𝒗; 𝒓 =
𝒎𝒗𝟐
𝑩𝒒𝒗
, r = 2.84 &times; 10-4m
ANTI-CLOCKWISE CIRCULAR MOTION TO THE RIGHT.
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