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Given: Remaining portion of the conductor has a uniform volume current density (J) [parallel to the axis]Current density =J\text { density }=JApplying Ampere's cirmital lawh \rightarrow height of hylindes\begin{array}{l} \oint_{0}^{2 \pi d} B_{1} \cdot d l=M_{0} J \times \pi d^{2} h \\\Rightarrow \quad B_{1} \times 2 \pi d=M_{0} \pi d^{2} h J . \\\Rightarrow \quad B_{1}=\frac{M_{0} d h J}{2}\end{array}Magnetic field at the center of second loop is zero.\therefore \text { Buet }=\frac{M \circ d h J}{2}As the expression for magnetic field contains J , d, and h. d and h is constant and as given in the question J is also uniform. Hence magnetic field in the hole is uniform. ...