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Question

A current i=2.5 A flows along the circular coil of three turns in anticlockwise direction. The equation of circle is given by x2+y2=9 cm2 (x & y are in cm). The magnetic field at point P (0,0,4 cm) is :
[consider x and y axis in the plane of paper and perpendicular outward direction as +z - axis ]

A
(36π×107 T) ^k
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B
(108π×107 T) ^k
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C
(9π5×107 T) (^k)
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D
(36π×107 T) (^k)
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Solution

The correct option is B (108π×107 T) ^k
From the equation of circle,
x2+y2=9 cm2
Comparing it with x2+y2=a2

We find that the circular coil is placed in xy plane with centre at origin (0,0) and radius R=3 cm.

From the shown figure it is clear that point P(x=0,y=0,z=4 cm) lies along the axis of coil line (i.e. zaxis)

Using the relation;

BP=(μ0iR2)n2(R2+x2)3/2

BP=(4π×107)×2.5×(3×102)2×32[(3×102)2+(4×102)2]3/2

BP=27×10π×10112[(25×104)]3/2

BP=108π×107 T

Also, from right-hand thumb rule direction of field at P is along +z- axis.

BP=(108π×107) ^k

Ans : (b)
Why this question ?
Tip: Just focus on the equation of circle, it will give the clue for plane of coil (XY), centre (0,0) & the radius (R). Thus, we can easily visualize the point P and apply the formula for magnetic field due to coil.

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