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a square loop of side 'a' is placed at distance 'd' from an infinite wire carrying I . the wire lies in the plane of the loop . work done in rotating the loop about axis AA' through an angle of 180degree.

a square loop of side 'a' is placed at distance 'd' from an infinite wire carrying I . the wire lies in the plane of the loop . work done in rotating the loop about axis AA' through an angle of 180degree.

Grade:Upto college level

1 Answers

Ramesh V
70 Points
11 years ago

I cant upload the image due to some problems but i can explain it

the wire is infinite ,and a square coil is present at side of lenght a and its center is at d distant from infinite loop.

The current I2 is passing in infinite wire upwards and I1 is current in square loop.

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The B ?eld is into the page everywhere on the right of the wire in the plane of the square loop. Its magnitude is given by B = µ0 I2 /2πR,
where R is the distance to the wire.
The above explanation shows the forces on the left and right hand side of the loop from this ?eld. These forces are obtained from
F = Il × B, and the magnitudes are
Fleft = I1 aB = I1 a* { µ0I2/ (2π(d − a/2)) }
Fright= I1 aB = I1 a* { µ0I2/ (2π(d + a/2))}

There is also an upward force (Ftop ) and (Ftop ), but they
are equal in magnitude and oppositely directed. Hence the
net force is the vector sum of the forces to the left and to
the right. The net force is to the left and has magnitude

Fleft − Fright = { µ0 I1 I2 a/2Pi}*{1/(d − a/2) - 1(d+a/2) }
                    =  { µ0 I1 I2 a/2Pi}*{ a2 / (d2- a2/4) }

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regards

Ramesh

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