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JEE Main Physics Class 12 Electromagnetic Induction Questions
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© examsnet.com
Question : 87
Total: 120
A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 Nm
‌
–
1
(see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of strip is 50 grams, its resistance 10Ω and air drag negligible, N will be close to :
[8 April 2019 I]
1000
50000
5000
10000
Validate
Solution:
Force on the strip when it is at stretched position x from mean position is
F
=
−
k
x
−
i
I
B
=
−
k
x
−
B
I
v
R
×
I
B
F
=
−
k
x
−
B
2
I
2
R
×
v
Above expression shows that it is case of damped oscillation, so its amplitude can be given by
⇒
A
=
A
0
e
−
b
t
2
m
⇒
A
0
e
=
A
0
e
−
b
t
2
m
[as per question
A
=
A
0
e
]
⇒
t
=
2
m
(
B
2
I
2
R
)
=
2
×
50
×
10
−
3
×
10
0.01
×
0.01
Given,
m
=
50
×
10
−
3
k
g
B
=
0.1
T
l
=
0.1
m
R
=
10
Ω
k
=
0.5
N
Time period,
T
=
2
Ï€
√
m
k
≃
2
s
so, required number of oscillations,
N
=
10000
2
=
5000
© examsnet.com
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