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JEE Advanced 2019 Paper 2
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© examsnet.com
Question : 9
Total: 54
A 10 cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R=1Ω as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1 T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is
x
×
10
−
3
A
, where the value of x is _____
[Assume the velocity of wire PQ remains constant (1 cm’s) after key S is closed.
Given:
e
−
1
=
0.37
, where e is base of the natural logarithm]
Your Answer:
Validate
Solution:
E
M
F
=
B
V
l
=
1
V
=
1
×
10
−
2
×
10
−
1
=
10
−
3
V
Time constant
=
L
R
=
1
×
10
−
3
1
=
1
m
.
s
e
c
i
=
i
o
(
1
−
e
−
t
5
)
i
=
10
−
3
1
(
1
−
e
−
1
)
=
10
−
3
×
0.63
=
0.63
m
A
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