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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 33
Total: 54
A small circular loop of area
A
and resistance
R
is fixed on a horizontal
x
y
-plane with the center of the loop always on the axis
∧
n
of a long solenoid. The solenoid has
m
turns per unit length and carries current
I
counterclockwiseas shown in the figure. The magnetic field due to the solenoid is in
∧
n
direction. List-I gives time dependences of
∧
n
in terms of a constant angular frequency
ω
. List-II gives the torques experienced by the circular loop at time
t
=
π
6
ω
, Let
α
=
A
2
µ
0
2
m
2
I
2
ω
2
R
.
List-I
List-II
(I)
1
√
2
(
sin
ω
t
∧
j
+
cos
ω
t
∧
k
)
(P)
0
(II)
1
√
2
(
sin
ω
t
∧
i
+
cos
ω
t
∧
j
)
(Q)
−
α
4
∧
i
(III)
1
√
2
(
sin
ω
t
∧
i
+
cos
ω
t
∧
k
)
(R)
3
α
4
∧
i
(IV)
1
√
2
(
cos
ω
t
∧
i
+
sin
ω
t
∧
k
)
(S)
α
4
∧
j
(T)
−
3
α
4
∧
i
Which one of the following options is correct?
I
→
Q
,
II
→
P
,
III
→
S
,
IV
→
T
I
→
S
,
II
→
T
,
III
→
Q
,
IV
→
P
I
→
Q
,
II
→
P
,
III
→
S
,
IV
→
R
I
→
T
,
II
→
Q
,
III
→
P
,
IV
→
R
Validate
Solution:
(I)
→
B
=
µ
0
mI
√
2
(
sin
ω
t
∧
j
+
cos
ω
t
∧
k
)
φ
=
→
B
⋅
→
A
=
µ
0
mI
√
2
cos
(
ω
t
)
⋅
A
ε
=
d
φ
dt
=
µ
0
mI
ω
A
√
2
sin
(
ω
t
)
i
=
ε
R
=
µ
0
mI
ω
A
√
2
R
sin
(
ω
t
)
→
M
=
i
→
A
=
iA
(
∧
k
)
=
µ
0
mI
ω
A
2
√
2
R
sin
(
ω
t
)
(
∧
k
)
→
τ
=
→
M
×
→
B
=
µ
0
m
2
I
2
ω
A
2
√
2
R
sin
2
(
ω
t
)
(
−
∧
i
)
=
−
(
α
4
)
∧
i
(II)
→
B
=
µ
0
mI
√
2
(
sin
ω
t
∧
i
+
cos
ω
t
∧
j
)
φ
=
0
,
ε
=
0
,
i
=
0
,
t
=
0
(III)
→
B
=
µ
0
mI
√
2
(
sin
ω
t
∧
i
+
cos
ω
t
∧
k
)
φ
=
→
B
⋅
→
A
=
µ
0
mI
√
2
⋅
cos
(
ω
t
)
⋅
A
ε
=
−
d
φ
d
t
=
µ
0
mI
ω
A
√
2
sin
(
ω
t
)
i
=
ε
R
=
µ
0
mI
ω
A
√
2
R
sin
(
ω
t
)
→
M
=
i
→
A
=
iA
(
∧
k
)
=
µ
0
mI
ω
A
2
√
2
R
sin
(
ω
t
)
(
∧
k
)
→
τ
=
→
M
×
→
B
=
µ
0
m
2
I
2
ω
A
2
2
R
sin
2
(
ω
t
)
(
+
∧
j
)
=
α
4
∧
j
(IV)
→
B
=
µ
0
mI
√
2
(
cos
ω
t
∧
j
+
sin
ω
t
∧
k
)
φ
=
→
B
⋅
→
A
=
µ
0
mI
√
2
⋅
sin
(
ω
t
)
⋅
A
ε
=
−
d
φ
dt
=
µ
0
mI
ω
A
√
2
cos
(
ω
t
)
i
=
ε
R
=
−
µ
0
mI
ω
A
√
2
R
cos
(
ω
t
)
→
M
=
i
→
A
=
iA
(
∧
k
)
=
−
µ
0
mI
ω
A
2
√
2
R
cos
(
ω
t
)
(
∧
k
)
→
τ
=
→
M
×
→
B
=
−
µ
0
m
2
I
2
ω
A
2
2
R
cos
2
(
ω
t
)
(
−
∧
i
)
=
α
⋅
cos
2
(
π
6
)
∧
i
=
3
α
4
∧
i
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