Exam Style Answers 26 Asal Physics CB

 

 

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.

 Coursebook answers Chapter 26 6

Exam-style questions 1

B 

[1]

2

The most obvious demonstration is to show that the secondary coil is made of insulated wire, so no current can flow from the core to the secondary coil. (Alternatively (Alternatively, if you arrange for a small gap in the core, perhaps a piece of paper, then there is still an induced e.m.f., even though paper is an insulator. The e.m.f.. will be reduced because the amount e.m.f

a

A magnetic flux linkage of 1 Wb exists if a coil of 1 turn and area 1 m 2  [1]

 

has a field of 1 T passing at right angles through it.  [1]

 

(Other areas and numbers of turns are possible.)

b

i

   ii ii

 

An  A n electrical current is induced because there is a change in the magnetic flux linking the secondary coil. This changing flux is caused by the changing current in the primary coil.   [1] a

Φ = BA 

[1]

 

= 20 × 10−3 × (5.0 × 10−2)2 = 5.0 × 10−5 Wb  ∆( N Φ  ) e.m.f. E  =    ∆t

[1]

b

100 × 5.0 ×10  

    4

= 

0.1 = 5.0 × 10−2 V

4

−

[1] [1] [1] [1]

 iii iii

e.m.f. correct and constant between e.m.f. 0 and 0.2 s and zero between 0.2 and 0.4 s  [1]

 

Negative e.m.f e.m.f.. of half half the value in ii  between 0.4 and 0.8 s  [1]

0.086

[1]

5

−

 

[1]

[1]

0

∆( N Φ  )

= =   ∆t e.m.f. E   Blv E = 5.0 × 10−5 × 40 × 300 

[1]

E = 0.60 V 

[1]

absolute uncertainty = 10/300 × 0.60 

[1]

absolute uncertainty = 0.02 V 

[1]

7

a

i

 ii ii

When there is no flux linkage linkage,, the flux is changing at the greatest rate and so the induced e.m.f. is a maximum.  

  [2]

0.4

0.8

0

Time/s

–0.043

[1]

When  W hen the flux linkage is a maximum, it is, instantaneously,, not changing and thus there instantaneously t here is no induced e.m.f.   [2]

1

 

 = 

e.m.f./V

(Therefore, V  = 0.60 ± 0.02 V) 5

0.6 ×1.2 ×10

  0.2 = 3.6 × 10−4 Wb s−1  ∆( N Φ  )  = 240 × 3.6 × 10−4  E  =  ∆t E  = 0.0864 ≈ 8.6 × 10−2 V  ∆t

of flux in the core is reduced if there is not a complete circuit of iron.)  [1]

3

∆Φ 

b

time taken = 

distance speed

 = 

0.02 0.5

 = 

4.0 × 10−2 s 

[1]

flux linkage = NBA NBA  =  150 × 0.30 × (0.02 × 0.02) 

[1]

flux linkage = 1.8 × 10−2 Wb 

[1]

The rate of change of magnetic flux is constant.  [1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

c

 

initial flux linkage = 0 and final flux linkage = 1.8 × 10−2 Wb  magnitude of induced e.m e.m.f. .f. = rate of change of magnetic flux linkage  1.8 ×10 10 2 − 0 E  =    4. 0 ×10 2 E  = 0.45 V (magnitude only) 

b [1] [1]

i

Φ = BA BA  = 50 × 10−3 × p × (0.1)2 

[1]

 

− − Φ = 1.57 × 10 3 ≈ 1.6 × 10 3 Wb 

[1]

 ii ii

change in flux linkage = ∆(  ∆(N  N Φ) =  −3 600 × 1.57 × 10  − 0 

[1]

−

    d

  e

−

[1]

 

[1]  iii iii

9

a

 

e.m.f. E  = 7.85 ≈ 7.9 V 

i

A change in magnetic flux causes an induced e.m.f.  [1]

 

The spokes are cutting magnetic lines of flux. or The circuit (containing a spoke and the connections) is sweeping out an area of magnetic flux.  [1]

 ii ii

Increase the strength of the magnetic field (magnetic flux density). This causes a greater magnetic flux linking the circuit and thus a greater rate of change of the magnetic flux.  [1]

 

Rotate the coil faster. Each change in magnetic flux occurs in a smaller time, and thus a greater rate of change of flux occurs.  [1]

i

area per second = pR2 f  = p × (0.15)2 × 5   [1]

 

area per second = 0.353 ≈ 0.35 m2 s−1  [1]

 ii  ii

E  = 

[1]

e.m.f. / V

0.45

0

8

0

0.04

0.08

Time / s

Correct axes and labels with e.m.f. constant between 0 and 0.04 s 

[1]

 

Zero e.m.f. between 0.04 and 0.08 s 

[1]

a

The magnitude of the e.m.f. e.m.f. induced is directly proportional to the rate of change [1]  of magnetic flux linkage.  of

[1]

Wb (magnitude [1] ∆( N Φ  ) 0.942 e.m.f. E  =  ∆t  =  0.12   [1] only) 

When the coil is completely within the field, the induced e.m.f. is zero.  [1] The reason for this is that there is no change in the magnetic flux linkage. 

∆(N Φ) = 0.942 ≈ 0.94

b

∆( N Φ  ) ∆t

 =

  ∆( BA) ∆t

[1]

 

= 0.353 × 5 × 10−3 

 

2

E  = 1.77 × 10−3 ≈ 1.8 × 10−3 V 

[1] [1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020