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
μ0I2πr(2r2−a24r2−a2) to the left
Let I be the current through the complete conductor.
Since I∝A, current through cavity will be I4.
For the complete conductor, B=μ0I2πr
For cavity 1, B1=μ0I42π(r−a2)
For cavity 2, B2=μ0I42π(r+a2)
Now, Bnet=B−(B1+B2)
Solving, we get Bnet=μ0I2πr(2r2−a24r2−a2)
The direction of the field will be towards left.