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Q.

Two long conductors are arranged as shown above to form overlapping cylinders, each of radius r, whose centers are separated by a distance d. Current of density J flows into the plane of the page along the shaded part of one conductor and an equal current flows out of the plane of the page along the shaded portion of the other, as shown. Point A is at the midpoint of the line joining the axes of the conductors.

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a

Magnetic field at A is (μ0/2π)4d2J/r, in the –y direction

b

Magnetic field inside the cavity is uniform

c

Magnetic field at A is (μ0/2π)d2/r, in the +y direction

d

Magnetic field at A is (μ0/2π)πdJ, in the +y direction

answer is A, D.

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Detailed Solution

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BA=BA1+BA2

BA12πd2=μ0Jπd22

BA1=μ0Jd4      (points perpendicular to OA, along +y)

BA22πd2=μ0Jπd22

BA2=μ0Jd4       (points perpendicular to O'A, along +y)

Net magnetic field at A

BA=BA1+BA2=μ0Jd4+μ0Jd4=μ0Jd2 (along +y) Option (a) is correct. Option (b) is wrong, Option (c) is wrong.

Consider an arbitrary point P which is at a distance x from point O and at a distance y from point O' .

BP12πx=μ0Jπx2

BP1=μ0Jx2

BP22πy=μ0Jπy2

BP2=μ0Jy2

x-components of BP1is BP1sinθ1           sinθ1=lx  
x-components of BP2 is BP2sinθ2          sinθ2=ly
x component of net magnetic field at P is μ0Jx2lxμ0Jy2ly=0
y component of net magnetic field at P is μ0Jx2d1x+μ0Jy2d2y=μ0Jd2    (along +y)
Magnetic field inside the cavity is uniform.    Option (d) is correct.

 

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Two long conductors are arranged as shown above to form overlapping cylinders, each of radius r, whose centers are separated by a distance d. Current of density J flows into the plane of the page along the shaded part of one conductor and an equal current flows out of the plane of the page along the shaded portion of the other, as shown. Point A is at the midpoint of the line joining the axes of the conductors.