33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Moving Charges and Magnetism
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47
µ 01 i ⎛ θ1 ⎞ µ 02 i ⎛ θ2
⎞
\ Net magnetic field, =
⎝
⎜
⎠
⎟ −
⎝
⎜
⎠
⎟
2R
2π
2R
2π
µ 01 i ⎛ 3π
⎞ µ 02 i ⎛ π ⎞
=
⎝
⎜
× ⎠
⎟ −
⎝
⎜
× ⎠
⎟
2R
2 2π
2R
2 2π
µ 0 1 2
⎡3i
i ⎤
= −
2R
⎣
⎢
4 4⎦
⎥
µ 0 1 1
⎡3i
3i
= −
2R
⎣
⎢
4 4
⎤
⎦
⎥ =
32. (b) : Let l be the length of the wire. Magnetic field
at the centre of the loop is
I
B = µ 0
2 R
∴ B = µπ 0 I
( l
l
= 2πR)
...(i)
µ 0nI
µ 0nI
µ
B′ = = or, 0n
2 πI
B′ =
...(ii)
2r
⎛
⎝ ⎜ l ⎞
l
2
⎠
⎟ 2 nπ
From eqns. (i) and (ii), we get B′ = n 2 B
33. (a) : Given situation is shown in the figure.
Parallel wires 1 and 3 are semiinfinite,
so magnetic field at O
due to them
0I
B1 = B3
=− µ 4πR k ^
Magnetic field at O due to semi-circular arc in
YZ-plane is given by B
0I
2 =− µ 4R i ^
Net magnetic
field
at point O is given by
B= B1 + B2 + B3
µ 0I
µ 0 µ 0
=− − −
4πR k ^ I
4R i ^ I
4πR k ^ µ =− 0I
^
π + ^
( i 2k)
4πR
34. (a) : Magnetic field induction
due to vertical loop at the centre O is
0I
B1
= µ
2R
It acts in horizontal direction.
Magnetic field induction due to horizontal loop at the
centre O is
0 I
B2
= µ 2
2R
It acts in vertically upward direction.
As B 1 and B 2 are perpendicular to each other, therefore
the resultant magnetic field induction at the centre O is
I I
Bnet = B + B = ⎛ 2
⎝ ⎜ ⎞
R ⎠
⎟ + ⎛ 2
⎝ ⎜
⎞
1 2 2 2 µ 0 µ 02
R ⎠
⎟
2 2
I
I
Bnet = µ 0 2
R
+ 2 = 5 µ 0
() 1 () 2
2
2R
35. (b) : The current flowing in the ring is
I = q f...(i)
The magnetic induction at the centre of the ring is
µ 0I
µ 0qf
B = = (Using(i))
2R
2R
36. (a) : The loop mentioned in the question must look
like one as shown in the figure.
0
Magnetic field at the
centre due to semicircular
loop lying in x-y plane,
i
Bxy = 1 ⎛ ⎝ ⎜ µ 0 ⎞
R ⎠
⎟ negative z
2 2
direction.
Similarly field due to loop in
i
x-z plane, Bxz = 1 ⎛ ⎝ ⎜ µ 0 ⎞
R ⎠
⎟ in negative y direction.
2 2
\ Magnitude of resultant magnetic field,
i i
B= Bxy
+ Bxz
= ⎛ 2
⎝ ⎜ ⎞
R ⎠
⎟ + ⎛ 2
2 2 µ 0
⎝ ⎜
µ 0 ⎞
R ⎠
⎟ = µ 0i
=
µ 0i
2
4 4 4R
2 2R
37. (*) : The magnetic field at the centre of the coil,
ni
B = µ 0 ; where r is the radius.
2r
E/R = i
\ R ∝ 2pr ⇒ R = cr, where c is a constant.
ni nE nE
\ In the first coil, B1
= µ 0 1 0 0
2r
= µ
1 2r1 cr
= µ
( 1)
2cr1 2
µ 0nE1
µ 0nE1
If r1 = 2r2,
B1
=
2c
2r2 2 =
( ) 2c⋅
4r2 2
0nE2
B2
= µ
2cr2 2
As B 1 will not be equal to B 2 unless E 1 is different from E 2 ,
E 1 and E 2 will not be the same.
It is wrong to ask what potential difference should be
applied across them. It should be perhaps the ratio of
potential differences.
In that case, B 1 = B 2 , E 1
E1
= E2 ⇒ E1 = 4E2
∴ = 4.
4
E2
*Question is not correct.
38. (c) : The magnetic field is produced by moving
i
electron in circular path B = µ 0
2r
q q
where i = = × v
t 2π r
µ 0qv
v
∴ B = ⇒ r ∝
2
4πr
B
39. (c) : The magnetic field B produced at the centre of a
circular coil due to current I flowing through this is given
NI
by B = µ 0
, N is number of turns and r is radius of the
2r
µ I
coil. Here B = 0
[ N = 1]
2r
Here, 2 × 2pr′ = 2pr \ r′ = r/2.
\ Magnetic field at the centre for two turns (N = 2) is
given by
µ 0 × 2I
µ 0 × 2I
4µ
0I
B′ = = = = 4B
.
2r′
2r
/ 2 2r
− 7
Ni
40. (a): B = µ 0 4π × 10 × 1000 × 01 .
=
= 628 × 10
−
.
2r
2×
0.1
4
T