33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Electromagnetic Induction
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\ Df = qR = (0.2 C) (10 W) = 2 weber
20. (b) : Induced emf is given by V = ∆ φ
Q
Q ∆t
current( i)
= ⇒ ∆ φ
∆ t ∆ t × 1
R
= ∆t
[where Q is total charge in time Dt]
∆φ
⇒ Q = R
21. (c) : q =∫ idt = 1
R ∫ dt = ⎛ − dφ⎞
1
⎝
⎜
dt ⎠
⎟
R ∫ dt = 1
ε
R ∫ d φ
Hence total charge induced in the conducting loop
depend upon the total change in magnetic flux.
As the emf or iR depends on rate of change of f, charge
induced depends on change of flux.
22. (a) : Electric heater works on the principle of Joule’s
heating effect.
23. (a) : Eddy currents are produced when a metal is
kept in a varying magnetic field.
24. (d) : Magnetic potential energy stored in an
inductor is given by
1 2 −3 1
−3 2
U = LI ⇒ 25 × 10 = × L× ( 60 × 10 )
2
2
6 −3
25 × 2× 10 × 10 500
L =
= = 13. 89 H
3600 36
25. (b) : Here, I = 2.5 A, L = 5 H
Magnetic flux linked with the coil is
f B = LI = (5 H)(2.5 A) = 12.5 Wb
26. (d) : | V | =− L dI
dt
|V| ∝ slope of I-t graph
27. (a) : Inducedemf , e=−L di
T
dt
For 0≤t
≤ ,
4
i-t graph is a straight
line with positive
constant slope.
di
∴ = constant
dt
⇒ e = –ve and constant
For T T
≤t
≤
4 2
di
i is constant ∴ =0 ⇒ e = 0
dt
For T 3T
≤t
≤
2 4
i-t graph is a straight line with negative constant slope.
di
∴ = constant
dt
⇒ e = +ve and constant
For 3 T ≤t ≤ T
4
di
i is zero ∴ =0 ⇒ e = 0
dt
vvv
From this analysis, the variation of induced emf with
time is as shown in the figure.
28. (a) : Net flux Nf = Li
Flux per turn = 4 × 10 –3 Wb, i = 2 A
L = N = × −3
φ 4 10 × 500
= 1 henry
i 2
29. (d) : Mutual inductance between coils is
M = K LL
−3 −3
1 2= 1 2× 10 × 8× 10 ( QK = 1)
= 4 × 10 –3 = 4 mH
30. (a) : I = t 2 e –t
| ε= | L dI
dI
here emf iszerowhen = 0
dt
dt
dI −t 2 −t −t 2 −t
= 2te − t e = 0;
2te = t e
dt
i.e., te –t (t – 2) = 0 ⇒ t ≠ ∞ and t ≠ 0 \ t = 2 sec
31. (b) : As , || ε=M dI
dt
= M d ( =
dt I 0sin ω t ) MI 0ωcosω
t
\ e max = 0.005 × 10 × 100p × 1 = 5p
N
Ni
32. (c) : L = φ = BA B=
ni
µ
φ µ 0
; ; 0 =
i
l
N
L = µ 2
0
A=
µ 0n 2 Al
l
where n is the number of turns per unit length L ∝ N 2
33. (b) : ε=− L di
dt
−3
−ε − 5×
10
L = = H=
5 mH
di ( 2−
3)
dt
34. (d) : Self inductance of a solenoid = m 0 n 2 Al
where n is the number of turns per length.
So self induction ∝ n 2
So inductance becomes 4 times when n is doubled.
35. (c) :
1
E= Li
2 1 −
= ( 100× 10 3 ) × 1 2 = 0. 05 J
2 2
36. (a) : | ε | = L di
dt
Given that, L = 40 × 10 –3 H, di = 11 A – 1 A = 10 A
and dt = 4 × 10 –3 s
−3
⎛ 10 ⎞
∴ | ε | = 40 × 10 ×
⎝
⎜ −
× ⎠
⎟ = 100 V
3
4 10
37. (c)
38. (d)
E
39. (c) : 0
I N BA ω
0 = =
R R
Given, N = 1, B = 10 –2 T, A = p(0.3) 2 m 2 , R = p 2 W
f = (200/60) and w = 2p(200/60)
Substituting these values and solving, we get
I 0 = 6 × 10 –3 A = 6 mA