RC
F = mrω2
In a magnetic field
Fields
F = BIl sin θ
Coulomb’s law
F = Bqv sin θ
Q Q
F = 1 2
4πε r2
Question
Physics
Edexcel2022
× 1.6 × 10 −19 × 197 × 1.6 × 10 −19
B r =
Question
Physics
CAIE2022
Qq
(cid:72)(cid:79)(cid:72)(cid:70)(cid:87)(cid:85)(cid:76)(cid:70)(cid:68)(cid:79) (cid:83)(cid:82)(cid:87)(cid:72)(cid:81)(cid:87)(cid:76)(cid:68)(cid:79) (cid:72)(cid:81)(cid:72)(cid:85)(cid:74)(cid:92) E =
P 4rf r
A device called a clutch can be used to connect a motor to a load. The diagram shows a
design called an eddy current clutch.
copper disc Side view
plastic disc load
magnet
motor
magnet motor
plastic disc copper disc
Several magnets are embedded in the plastic disc and it is rotated by the motor.
(a) (i) Explain why a current is induced in the copper disc when the motor is switched on.
(2)
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(ii) Explain, using Lenz’s law, why the copper disc rotates.
(3)
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Question
Physics
Edexcel2022
RC
F = mrω2
In a magnetic field
Fields
F = BIl sin θ
Coulomb’s law
F = Bqv sin θ
Q Q
F = 1 2
4πε r2
Question
Physics
Edexcel2022
Faraday’s and Lenz’s laws
Electric field strength
−d(Nϕ)
E =
F dt
E =
Q
Root‑mean‑square values
Q
E =
4πε r2 V
*P67096A02732*
PMT
(c) The coil within a very sensitive moving coil ammeter can be damaged when the
ammeter is transported. The two ends of the coil are connected together when the
ammeter is transported. This reduces the movement of the coil and makes it less
likely to be damaged.
A student suggests that this is due to Faraday’s law and Lenz’s law.
Explain how these laws apply to this situation.
(4)
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(Total for Question 17 = 13 marks)
TOTAL FOR PAPER = 90 MARKS
Question
Physics
Edexcel2021
RC
F = mrω2
In a magnetic field
Fields
F = BIl sin θ
Coulomb’s law
F = Bqv sin θ
Q Q
F = 1 2
4πε r2
Question
Physics
Edexcel2021
Faraday’s and Lenz’s laws
Electric field strength
−d(Nϕ)
E =
F dt
E =
Q
Root-mean-square values
Q
E =
4πε r2 V
Question
Physics
Edexcel2021
The diagrams show the plan view and side view of a moving coil ammeter.
Plan view Side view
scale
coil
I
B
pointer
soft iron cylinder soft iron
pivot
N l S
cylinder
B
w
N S
pivot
magnet magnet
moving coil
The fixed soft iron cylinder and magnets produce a uniform magnetic field of magnetic
flux density B. The coil is able to rotate within this magnetic field. The coil has width
w and length l. There is a current I in the coil in the direction shown in the side view
diagram.
(a) (i) Explain which way the coil will rotate.
(2)
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Question
Physics
Edexcel2020
RC
F = mrω2
In a magnetic field
Fields
F = BIl sin θ
Coulomb’s law
F = Bqv sin θ
Q Q
F = 1 2
4πε r2
Question
Physics
Edexcel2020
Faraday’s and Lenz’s laws
Electric field strength
−d(Nϕ)
E =
F dt
E =
Q
Root-mean-square values
Q
E =
4πε r2 V
Question
Physics
Edexcel2020
Hybrid electric vehicles (HEV) use the same device both as a generator to charge the car
battery and as an electric motor to support the propulsion system. A simplified diagram
of the device is shown. The coil can rotate freely around the axis.
coil
magnet magnet
S N
axis
*(a) Describe how the device can be used as both a generator and an electric motor.
(6)
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Question
Physics
CAIE2016
Data
speed of light in free space c = 3.00 (cid:117) 108 ms–1
permeability of free space (cid:80) = 4(cid:83) (cid:117) 10–7 Hm–1
Question
Physics
CAIE2016
2
BI
Hall voltage V =
H ntq
alternating current/voltage x = x sin (cid:90)(cid:3)(cid:3)t