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Electromagnetic Induction
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Section:
Physics
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© examsnet.com
Question : 4
Total: 18
A thin conducting rod MN of mass
20
‌
gm
, length
25
‌
cm
and resistance
10
Ω
is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field
B
0
=
4
T
directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time
t
=
0
and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.
[Given: The acceleration due to gravity
g
=
10
m
s
−
2
and
e
−
1
=
0.4
]
List-I
List-II
(P) At
t
=
0.2
s
, the magnitude of the induced emf in Volt
(1) 0.07
(Q) At
t
=
0.2
s
, the magnitude of the magnetic force in Newton
(2) 0.14
(R) At
t
=
0.2
s
, the power dissipated as heat in Watt
(3) 1.20
(S) The magnitude of terminal velocity of the rod in
m
s
−
1
(4) 0.12
(5) 2.00
[JEE Adv 2023 P1]
P
→
5
,
Q
→
2
,
R
→
3
,
S
→
1
P
→
3
,
Q
→
1
,
R
→
4
,
S
→
5
P
→
4
,
Q
→
3
,
R
→
1
,
S
→
2
P
→
3
,
Q
→
4
,
R
→
2
,
S
→
5
Validate
Solution:
Induced emf
ε
=
B
â„“
v
⇒
Induced current
i
=
‌
ε
R
=
‌
B
â„“
V
R
⇒
m
g
−
i
â„“
B
=
m
a
[Applying
2
‌
nd
‌
law]
⇒
‌
‌
m
g
−
‌
B
2
â„“
2
v
R
=
m
‌
d
v
d
t
⇒
‌
d
v
m
g
−
‌
B
2
â„“
2
v
R
=
‌
d
t
m
‌
‌
⇒
‌
‌
‌
ln
[
m
g
−
B
2
â„“
2
v
R
]
0
v
−
B
2
â„“
2
R
=
‌
t
m
⇒
‌
m
g
−
B
2
â„“
2
v
R
m
g
=
e
‌
−
B
2
â„“
2
m
R
t
⇒
‌
‌
v
=
2
[
1
−
e
−
5
t
]
⇒
At
t
=
0.2
s
,
v
=
2
[
1
−
‌
1
e
]
⇒
‌
‌
ε
=
B
â„“
×
2
[
1
−
‌
1
e
]
=
1.2
volts
and magnetic force
=
i
â„“
B
=
0.12
N
and power dissipated
=
0.144
W
also, Terminal velocity
=
2
m
∕
s
⇒
Correct match is (D)
© examsnet.com
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