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Question

An infinitely long wire carries a current I flowing towards positive z - axis. Consider a circle in xy plane with centre at (2m,0,0) and radius 1 m as shown in figure.

The maximum value of path integral B.dl of the magnetic field B due to current carrying wire between two points P & Q lying along the circular path is

A
μ0I12
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B
μ0I8
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C
μ0I6
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D
Zero
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Solution

The correct option is C μ0I6
Two point P and Q will be equidistant from the current carrying wire placed at origin due to the symmetry.

Also, the distance of any point on the arc PQ from wire position O will be equal.

The magnetic field due to wire will be along clockwise direction on the arc PQ.

Magnetic field due to wire,

B=μ0I2πd=μ0I2π(OP)

Now B.dl=Bdlcos0

or, B.dl=Bdl

Here dl is the length of arc PQ

PQ=(OP)(2θ)

Thus, B.dl=μ0I2π(OP)×(OP) (2θ) .........(1)

For the maximum value of B.dl, according to equation, the value of 2θ should be maximum. It can happen only if P and Q are at position such that OP & OQ are tangents to the circle.


sinθ=PCOC

sinθ=12

θ=30

Hence, 2θ=602θ=π3

The maximum value of B.dl is given by,

(B.dl)max=μ0I2π×π3=μ0I6

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