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AIPMT 2008 Physics and Chemistry Paper
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© examsnet.com
Question : 31
Total: 100
A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are
F
1
,
F
2
and
F
3
respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is :
√
(
F
3
−
F
1
)
2
−
F
2
2
F
3
−
F
1
−
F
2
F
3
−
F
1
−
F
2
√
(
F
3
−
F
1
)
2
+
F
2
2
Validate
Solution:
Total force on the current carrying closed loop should be zero, if placed in uniform magnetic field.
F
horizontal
=
(
F
3
−
F
1
)
F
vertical
=
F
2
Resultant of
→
F
1
,
→
F
2
and
→
F
3
is
→
F
where
F
=
√
(
F
3
−
F
1
)
2
+
F
2
2
Since total force
=
0
, hence force on
QP
is equal to
→
F
in magnitude but opposite direction.
F
QP
=
√
(
F
3
−
F
1
)
2
+
F
2
2
© examsnet.com
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