Electromagnetic Induction- 2 Practice Questions - DPP for NEET

 Page 1


1. (a). e = – L 
dI
dt
 = – 10 × 10
–3
 
1.0
0.01
 = –1V olt.
\ | e | = 1 volt.
2. (a). e = – L 
dI
dt
 = – L 
d
dt
 [3t
2
 + 2t]
= – L [6t + 2] = –10 × 10
–3
 [6t + 2]
(e)
at t = 2 
= – 10 × 10
–3
 (6 × 2 + 2)
= – 10 × 10
–3
 (14) = – 0.14 V olt
 |e| = 0.14 volt.
3. (a). The mutual inductance between two coils depends on
their degree of flux linkage, i.e., the fraction of flux linked
with one coil which is also linked to the other coil. Here,
the two coils in arrangement (a) are placed with their planes
parallel. This will allow maximum flux linkage.
4. (b). e
2
 = – M
1
dI
dt
= 
3
0 10
3 10
-
-
-
´
  = 3 × 10
4
 volt = 30 kV . .
5. (d). L
2
 and L
3
 are in parallel. Thus their combination gives
L' = 
23
23
LL
LL +
 = 0.25 H
The L' and L
1
 are in series, thus the equivalent inductance
is L = L
1
 + L' = 0.75 + 0.25 = 1H
6. (a). We use e = – L DI/Dt to determine the value of induced
emf. (i) DI = (7 – 0) = 7A,
Dt = (2 – 0) ms = 2ms
Thus e = – 4.6 × 
3
7
2 10
-
´
 = – 16.1 × 10
3
 volt
(ii) DI = 5 – 7 = – 2A,
 dt = (5 – 2) ms = 3 ms
Thus e = – 4.6 × 
3
( 2)
3 10
-
-
´
 = 3.07 × 10
3
 V
7. (b). I = I
0
 (1 – e
– t/t
)
Where I = 
1
2
 I
0
 and t = L/R
Thus    
1
2
 I
0
 = I
0
 (1 – e
 – t/t 
)
or  
1
2
 = e
 – t/t 
 or 2 = e
+ t/t
  or log 2 = t/t
Thus  t = t log
e
 2 = 
3
50 10
0.025
-
´
 × 0.693 = 1.385
8. (a). V
S
 = I
S
 Z
S
 Þ 22 = I
S
 × 220
\ I
S
 = 0.1A
S P
PS
V I
VI
=
P
I 22
220 0.1
= Þ I
P
 = 0.01 A.
9. (c). V
p
 = 220V , I
p
 = 5S, V
s
 = 2200V
P
s
 = 
p
P
2
, I
s
 = ?
Q  V
s
I
s
 = 
pp
VI
2
After putting the given  value you will find
I
s
 = 0.25 A.
10. (d).  During decary of current
i = 
R
L
0
t
ie
-
 = 
R
L
E
R
t
e
 = 
3
3
100 10
100 10
100
100
e
-
-
´
´
 = 
1
A.
e
11. (c). In a generator e.m.f. is induced according as Lenz’s
rule.
The minus sign indicates that the direction of the induced
e.m.f. is such as to oppose the change in current.
12. (a). 'Immediately' after pressing the switch S, the current in
the coil L, due to its self-induction will be zero, that is
i
2
 = 0.
The current will only be found in the resistance R
1
 and this
will be the total current in the circuit.
\  i = i
1
  = 
2
E 10 volt
R 5.0 volt
= = 2.0 ampere.
13. (a). (i) In series the same current i will be induced in both
the inductors and the total magnetic-flux linked with them
will be equal to the sum of the fluxes linked with them
individualy , that is,
F = L
1
i + L
2
i.
If the equivalent inductance be L, then F = Li.
\ Li = L
1
i + L
2
i or  L = L
1
 + L
2
.
(ii) In parallel, let the induced currents in the two coils be i
1
and i
2.
 Then the total induced current is
i = i
1
 + i
2
    \ 
12
di di di
dt dt dt
=+
In parallel, the induced e.m.f. across each coil will be the
same.
Hence   e = – L
1
1
di
dt
 = –L
2
2
di
dt
.
If the equivalent inductance be L, then e = – L
di
dt
.
\  
e di
L dt
=- = – 
12
di di
dt dt
æö
+
ç÷
èø
 = 
12
ee
LL
+
Page 2


1. (a). e = – L 
dI
dt
 = – 10 × 10
–3
 
1.0
0.01
 = –1V olt.
\ | e | = 1 volt.
2. (a). e = – L 
dI
dt
 = – L 
d
dt
 [3t
2
 + 2t]
= – L [6t + 2] = –10 × 10
–3
 [6t + 2]
(e)
at t = 2 
= – 10 × 10
–3
 (6 × 2 + 2)
= – 10 × 10
–3
 (14) = – 0.14 V olt
 |e| = 0.14 volt.
3. (a). The mutual inductance between two coils depends on
their degree of flux linkage, i.e., the fraction of flux linked
with one coil which is also linked to the other coil. Here,
the two coils in arrangement (a) are placed with their planes
parallel. This will allow maximum flux linkage.
4. (b). e
2
 = – M
1
dI
dt
= 
3
0 10
3 10
-
-
-
´
  = 3 × 10
4
 volt = 30 kV . .
5. (d). L
2
 and L
3
 are in parallel. Thus their combination gives
L' = 
23
23
LL
LL +
 = 0.25 H
The L' and L
1
 are in series, thus the equivalent inductance
is L = L
1
 + L' = 0.75 + 0.25 = 1H
6. (a). We use e = – L DI/Dt to determine the value of induced
emf. (i) DI = (7 – 0) = 7A,
Dt = (2 – 0) ms = 2ms
Thus e = – 4.6 × 
3
7
2 10
-
´
 = – 16.1 × 10
3
 volt
(ii) DI = 5 – 7 = – 2A,
 dt = (5 – 2) ms = 3 ms
Thus e = – 4.6 × 
3
( 2)
3 10
-
-
´
 = 3.07 × 10
3
 V
7. (b). I = I
0
 (1 – e
– t/t
)
Where I = 
1
2
 I
0
 and t = L/R
Thus    
1
2
 I
0
 = I
0
 (1 – e
 – t/t 
)
or  
1
2
 = e
 – t/t 
 or 2 = e
+ t/t
  or log 2 = t/t
Thus  t = t log
e
 2 = 
3
50 10
0.025
-
´
 × 0.693 = 1.385
8. (a). V
S
 = I
S
 Z
S
 Þ 22 = I
S
 × 220
\ I
S
 = 0.1A
S P
PS
V I
VI
=
P
I 22
220 0.1
= Þ I
P
 = 0.01 A.
9. (c). V
p
 = 220V , I
p
 = 5S, V
s
 = 2200V
P
s
 = 
p
P
2
, I
s
 = ?
Q  V
s
I
s
 = 
pp
VI
2
After putting the given  value you will find
I
s
 = 0.25 A.
10. (d).  During decary of current
i = 
R
L
0
t
ie
-
 = 
R
L
E
R
t
e
 = 
3
3
100 10
100 10
100
100
e
-
-
´
´
 = 
1
A.
e
11. (c). In a generator e.m.f. is induced according as Lenz’s
rule.
The minus sign indicates that the direction of the induced
e.m.f. is such as to oppose the change in current.
12. (a). 'Immediately' after pressing the switch S, the current in
the coil L, due to its self-induction will be zero, that is
i
2
 = 0.
The current will only be found in the resistance R
1
 and this
will be the total current in the circuit.
\  i = i
1
  = 
2
E 10 volt
R 5.0 volt
= = 2.0 ampere.
13. (a). (i) In series the same current i will be induced in both
the inductors and the total magnetic-flux linked with them
will be equal to the sum of the fluxes linked with them
individualy , that is,
F = L
1
i + L
2
i.
If the equivalent inductance be L, then F = Li.
\ Li = L
1
i + L
2
i or  L = L
1
 + L
2
.
(ii) In parallel, let the induced currents in the two coils be i
1
and i
2.
 Then the total induced current is
i = i
1
 + i
2
    \ 
12
di di di
dt dt dt
=+
In parallel, the induced e.m.f. across each coil will be the
same.
Hence   e = – L
1
1
di
dt
 = –L
2
2
di
dt
.
If the equivalent inductance be L, then e = – L
di
dt
.
\  
e di
L dt
=- = – 
12
di di
dt dt
æö
+
ç÷
èø
 = 
12
ee
LL
+
DPP/ P 45
126
or  
12
1 11
L LL
=+  or L = 
12
12
LL
LL +
.
14. (d). If we try to find field of the small coil and then calculate
flux through long solenoid, the problem becomes very
difficult. So we use the following fact about mutual
inductance.
M
21
 = M
12
, 
21
12
II
ff
=
Thus if I current flows in long solenoid, then flux f through
small coil is the same as the flux f
2
 that is obtained when I
current flows through the small coil. Therefore,
f
2 
= f
1
 = (Field at small coil) × (area) × (turns)
= 
2
0
2
N
I
æö
m
ç÷
èø l
 (AN
1
) = 
0 12
2
N N AI m
l
15. (a). The induced e.m.f. is
e = – M
i
t
D
D
or M  = – 
e
i/t DD
Here e = 1500 volt.
\ M = – 
1500
(0 3.0) / 0.001 -
 = 
1500 0.001
3.0
´
 = 0.5 henry . .
16. (c).  
S
E = 
I
M
p
d
dt
 = 
( )
0
I sin
M
dt
dt
w
= MI
0
w |cos wt|
Þ Crest value = MI
0
w
= 1.5 ´ 1 × 2p × 50
= 471 V
17. (a) ()
2 23
11
U Li 100 10 10 5J
22
-
= =´ ´ ´=
18. (c) Transformation ratio 
ss
s
p
VV 5
k V 100V
V 3 60
= Þ = Þ=
19. (b)
pp
s
p
sp
ii
N 1
i 0.04 A
i N 4 100
= Þ = Þ=
20. (a)
Output 80 20 120
Input 100 1000 i
l
´
h= Þ=
´
20 120 100
i 3A.
1000 80
´´
Þ==
´
l
21. ()
22. (b)
23. (a)
24. (b)
25. (a) 26.  (a) 27. (b)
Consider a strip at a distance x from the wire of thickness dx.
Magnetic flux associated with this strip
0
Ia
Badx dx
2x
m
Df==
p
E D
C
dx
F
A B
I
l
x
a 2a
00
a
Ia Ia dx dx 2a
1n
2 x x2
++
+
éù
mm+ æö
êú f= +=
ç÷
èø pp
êú
ëû
òò
ll
ll
l
l
0
(primary)
a 2a
M M 1n
I2
m f+ æö
= Þ=
ç÷
p
èø
l
l
dI
eM
dt
=-
00
0
Ia 2a
e MI 1n
2
m + æö
= - =-
ç÷
èø
p
l
l
Heat produced 
( )
0
2
2a
2
0
2
1n I at
e
t
R8
l
l
m
+
p
éù
êú
ëû
==
l
28. (a) Hysteresis loss in the core of transformer is directly
proportional to the hysteresis loop area of the core
material. Since soft iron has narrow hysteresis loop
area, that is why soft iron core is used in the transformer.
29. ( d) Efficiency of electric motor is maximum when the back
emf set up in the armature is half the value of the applied
battery emf.
30. (a) Transformer works on ac only , ac changes in magnitude
as well as in direction.