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JEE Main 7 Jan 2020 Shift 1 Solved Paper
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© examsnet.com
Question : 3
Total: 75
A long solenoid of radius R carries a time (t)-dependent current
I
(
t
)
=
I
0
t
(
1
−
t
)
. A ring of radius 2R is placed coaxially near its middle. During the time interval
0
≤
t
≤
1
, the induced current
(
I
R
)
and the induced EMF
(
V
R
)
in the ring change as :
[7-Jan-2020 Shift 1]
At t = 0.5 direction of
I
R
reverses and
V
R
is zero
Direction of
I
R
remains unchanged and
V
R
is zero at t = 0.25
Direction of
I
R
remains unchanged and
V
R
is maximum at t = 0.5
At t = 0.25 direction of
I
R
reverses and
V
R
is maximum
Validate
Solution:
Magnetic flux (ϕ) through ring is ϕ
=
π
(
R
)
2
.
B
ϕ
=
(
π
R
2
)
(
µ
0
n
I
)
=
(
π
R
2
µ
0
n
I
0
)
(
t
−
t
2
)
Induced e.m.f. of
V
R
=
−
d
ϕ
d
t
=
(
π
R
2
µ
0
n
I
0
)
(
2
t
−
1
)
and induced current
I
R
=
π
R
2
µ
0
n
I
0
(
2
t
−
1
)
R
R
(
R
R
→ Resistance of Ring)
Clearly
V
R
and
I
R
are zero a
t
=
1
2
=
0.5
sec
© examsnet.com
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