33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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48 NEET-AIPMT Chapterwise Topicwise Solutions Physics
−7
Ni
41. (b) : B = µ 0( ) 4π
× 10 × 50×
2
=
−4
= 1. 256 × 10 T
2r
2×
0.
5
42. (b) : Magnetic field at the centre of the coil,
NI
B = µ 0
2π
a
Let l be the length of the wire, then
µ 0 1× I µ 0I
B1
= ⋅ =
2π
l / 2π
l
and B µ 0 2 × I 4µ
0I
2 = ⋅ =
2π
l / 4π
l
B1
1
Therefore, = or, B1: B2
=1:
4
B2
4
43. (d) : Magnetic field due to long solid cylindrical
conductor of radius R,
2
Id
(i) For d < R, I′=
2
R
2
µ 0Id
µ 0Id
∫ Bdl . = µ 0I′ ⇒ B( 2πd)
= ⇒ B =
2
2
R
2πR
\ B ∝ d
(ii) For d = R, B = m 0 I/2pR (maximum)
(iii) For d > R, B = m 0 I/ 2pd ⇒ B ∝ 1/d
44. (a) : Magnetic field at a point inside the wire at
⎛ a ⎞
distance r =
⎝
⎜
⎠
⎟ from the axis of wire is
2
µ 0I
B
a r µ 0I
a µ 0I
= = × =
2 2
2π
2πa
2 4πa
Magnetic field at a point outside the wire at a distance
r(= 2a) from the axis of wire is
µ 0I
µ 0I
1 µ 0I
B
B′ = = × = ∴ = 1
2πr
2π
2a
4πa
B′
45. (d) : The magnetic field at the point P, at a
perpendicular distance d from O in a direction
perpendicular to the plane ABCD due to currents
through AOB and COD are perpendicular to each other.
Hence,
B= ( B + B ) /
1 2 2 212
⎡
= ⎛ 2
⎝ ⎜ ⎞
⎠
⎟ + ⎛ 2
µ
⎝ ⎜
µ ⎞
⎤
12 /
⎢ 0 2I1
0 2I2
⎠
⎟ ⎥
⎣ 4π
d 4π
d ⎦
µ 0
= 1 2 + 2 212
2πd ( I I ) /
i
46. (c) : B = µ 0 2 2 i
r
− µ 0 2 1
π π r = µ 0 4
(
π r i 2 − i 1)
4 ( / 2) 4 ( / 2)
4
µ 0 4
µ 0
= ( − =−
4π
5 25 . 50 . ) .
2π
Negative sign shows that B is acting inwards i.e., into the
plane.
47. (d) : Diameter of first wire (d 1 ) = 0.5 mm;
Current in first wire (I 1 ) = 1 A; Diameter of second wire
(d 2 ) = 1 mm and current in second wire (I 2 ) = 1 A
Strength of magnetic field due to current flowing in a
conductor, (B) = µ 0 2
× I
4π
a
or B ∝ I
Since the current in both the wires is same, therefore
there is no change in the strength of the magnetic field.
∫
48. (b) : Use Ampère’s law Bdl . = µ 0 i enclosed .
Outside : i enclosed ≠ 0 (some value) ⇒ B ≠ 0
Inside = i enclosed = 0 ⇒ B = 0
49. (b) : The direction of the magnetic field, due to
current, is given by the right-hand rule. At axis AB, the
components of magnetic field will cancel each other and
the resultant magnetic field will be zero.
i
50. (a) : B = µ 0
B
πr
∝ 1
or
2
r
When r is doubled, the magnetic field becomes halved
i.e.,now the magnetic field will be 0.2 T.
51. (a) : B ∝ 1/r, for given current.
As the distance is increased to three times, the magnetic
induction reduces to one third.
1 −3 −4
HenceB = × 10 T= 333 . × 10 T
3
52. (a) : Here, l = 50 cm, N = 100, i = 2.5 A
Magnetic field inside the solenoid,
B = m 0 ni = µ 0 NI
l
−7
4π× 10 × 100 × 25 .
B = = 6.28 × 10 –4 T
05 .
53. (a) : For a toroid magnetic field, B = m 0 ni
Where, n = number of turns per unit length = N 2πr
Now, B 1 µ 0ni
1 n1
N1
2πr2
= = = ×
B µ ni n 2πr
N
⇒
⇒
B
B
B
B
1
2
1
2
2
0 2
2
200
=
2π
× 40×
10
−2
1
= ⇒ B1 : B2
= 1:
1
1
1
2
2π
× 20×
10
×
100
54. (b) : Magnetic field induction at point inside a
long solenoid l, having n turns per unit length carrying
current i is given by
B = m 0 ni
If i → doubled, n → halved then B → remains same.
55. (c) : Force between wires A and B = force between
wires B and C
µ 0I 2 l
∴ FBC
= FAB
=
2 πd
As, F
F
AB ⊥ BC
net force on wire B,
2
oI l
Fnet = 2 FBC
= 2 µ F
or net oI
= µ 2
2πd
l 2 πd
56. (c) : Force on arm AB due to current in conductor
XY is
IiL Ii
F1
= µ 0 2 0
4π
L 2
= µ
( / ) π
acting towards XY in the plane of loop.
−2