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Question

A long, straight metal has a long hole of radius a drilled parallel to the rod axis with cross-sectional view as shown. If the rod carries a current I, the value of the magnetic field on the axis of the rod is:
1026592_fe82758a3c674f1dac856b74055bc394.png

A
B=μ0Ia22πc(b2a2)
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B
B=μ0I2πc
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C
B=μ0Ia24πc(b2a2)
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D
B=zero
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Solution

The correct option is A B=μ0Ia22πc(b2a2)
Suppose that cavity contains same the current and same -ve current which is overall zero.

as J=Iπ(b2a2),

current in cavity I1=Iπa2π(b2a2)
I1=Ia2(b2a2)
Then now we will calculate the magnetic field due to a cylinder of 'b' radius with current (+ve) and another cylinder of radius a with current (-ve).(Here positive and negative only denote the direction of flowing current w.r.t each other.)

B0=Bcylinder+BCavity

=μ0I12πd1+μ0I22πd2

d1=0, d2=c
On keeping value of d1and d2

BO=0+μIa22π(b2a2)1C=μIa22πC(b2a2)

B=μ0Ia22πC(b2a2)

Option A is correct.

1450500_1026592_ans_ffe6d23295924631bf50c9eb4dabc95b.png

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