A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be:
2μ0Ii3π
μ0Ii2π
2μ0IiL3π
μ0IiL2π
A
2μ0Ii3π
B
μ0Ii2π
C
μ0IiL2π
D
2μ0IiL3π
Open in App
Solution
Verified by Toppr
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=μ0I2πr
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 F1=(μ0I2πL/2)iL=μ0Iiπ and
the force on CD is F2=(μ0I2π(L+L/2))iL=μ0Ii3π
As the current directions on AB and CD are opposite so the forces will be opposite.
Thus, net forec F=F1−F2=μ0Iiπ−μ0Ii3π=2μ0Ii3π
Was this answer helpful?
42
Similar Questions
Q1
A square loop ABCD, carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be
View Solution
Q2
A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be:
View Solution
Q3
A square loop ABCD, carrying a current I1, is placed near and coplanar with a long straight conductor XY carrying a current I1, as shown in figure. The net force on the loop will be
View Solution
Q4
A square loop ABCD having side l carrying a current i, is placed at a distance l/2; coplanar and parallel with a long straight conductor XY carrying a current I, the net force on the loop will be
View Solution
Q5
A square loop ABCD having side l carrying a current i, is placed at a distance l/2; coplanar and parallel with a long straight conductor XY carrying a current I, the net force on the loop will be