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JEE Main 16 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 17
Total: 90
The magnetic field in a region is given by
B
=
B
0
(
x
a
)
∧
k
. A square loop of side
d
is placed with its edges along the
X
and
Y
-axes. The loop is moved with a constant velocity
v
=
v
0
∧
i
. The emf induced in the loop is
[16 Mar 2021 Shift 2]
B
0
v
0
2
d
2
a
B
0
v
0
d
2
a
B
0
v
0
d
2
a
B
0
v
0
d
2
2
a
Validate
Solution:
The given situation can be shown as
From the above figure, it can be seen that emf induced in side
D
C
and
A
B
will be zero. It is because
v
is parallel to length of conductor along
X
-axis.
∴
Net emf induced
=
E
B
C
−
E
A
D
=
B
0
(
x
+
d
)
v
0
d
a
−
B
0
x
v
0
d
a
=
B
0
v
0
d
2
a
© examsnet.com
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