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JEE Main Physics Class 12 Electromagnetic Induction Questions
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© examsnet.com
Question : 60
Total: 120
A circular conducting coil of radius
1
m
is being heated by the change of magnetic field
→
B
passing perpendicular to the plane in which the coil is laid. The resistance of the coil is
2
µ
Ω
. The magnetic field is slowly switched off such that its magnitude changes in time as
B
=
4
π
×
10
−
3
T
(
1
−
t
100
)
The energy dissipated by the coil before the magnetic field is switched off completely is
E
=
─
─
─
─
m
J
[25 Jul 2021 Shift 1]
Your Answer:
Validate
Solution:
ϕ
=
→
B
.
→
S
ϕ
=
4
π
×
10
−
3
(
1
−
t
100
)
.
π
R
2
ϕ
=
4
×
10
−
3
×
(
1
)
2
(
1
−
t
100
)
ε
=
−
d
ϕ
d
t
ε
=
−
d
d
t
(
4
×
10
−
3
(
1
−
t
100
)
)
ε
=
4
×
10
−
3
(
1
100
)
=
4
×
10
−
5
V
When
B
=
0
1
−
t
100
=
0
t
=
100
s
e
c
Heat
=
ε
2
R
t
Heat
=
(
4
×
10
−
5
)
2
2
×
10
−
6
×
100
J
Heat
=
16
×
10
−
10
×
100
2
×
10
−
6
J
Heat
=
0.08
J
Heat
=
80
m
J
© examsnet.com
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