A wire YY' is parallel to a square loop ABCD with AD || YY' . if current flow Y to Y' , and in clockwise direction in the square loop . state what you will observe and why?
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Answer:
Here the sides AB and CD will contribute the force but the sides AD and BC will not because they are perpendicular to wire XY.
We know that the magnetic field at distance r from long wire is B=
2πr
μ
0
I
and also force on a current (i) flow through line section of legth L due to B is F=BiL
Using these, the force on AB is F
1
=(
2πL/2
μ
0
I
)iL=
π
μ
0
Ii
and
the force on CD is F
2
=(
2π(L+L/2)
μ
0
I
)iL=
3π
μ
0
Ii
As the current directions on AB and CD are opposite so the forces will be opposite.
Thus, net forec F=F
1
−F
2
=
π
μ
0
Ii
−
3π
μ
0
Ii
=
3π
2μ
0
Ii
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