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Question

A long cylindrical conductor of radius 'a' has two cylindrical cavities each of diameter 'a' through its entire length as shown in the figure. A current I is directed out of the page and is uniform throughout the cross-section of the conductor. The magnetic field at point P1 is


A

μ0I2πr(2r2+a24r2a2) to the right

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B

μ0I2π(r2+a2r2a2) to the right

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C

μ0Iπr(2r2+a22r2a2) to the left

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D

μ0I2πr(2r2a24r2a2) to the left

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Solution

The correct option is D

μ0I2πr(2r2a24r2a2) to the left


Let I be the current through the complete conductor.
Since IA, current through cavity will be I4.
For the complete conductor, B=μ0I2πr
For cavity 1, B1=μ0I42π(ra2)
For cavity 2, B2=μ0I42π(r+a2)
Now, Bnet=B(B1+B2)
Solving, we get Bnet=μ0I2πr(2r2a24r2a2)
The direction of the field will be towards left.


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