Frequently, on a 32-bit CPU, each page-table entry is 4 bytes long, but that size can vary as well. A 32-bit entry can point to one of 2^32 physical page frames. If frame size is 4 KB (2^12), then a system with 4-byte entries can address 2^44 bytes (or 16 TB) of physical memory. We should note here that the size of physical memory in a paged memory system is different from the maximum logical size of a process.
How does paging make logical memory space more than physical memory space? Isn't total number of frames in 32-bit CPU equal to 2^(32-12)=2^20 number of frames rather than 2^32 number of frames? If so, isn't a system with 4-byte entry capable of addressing (2^20)*(2^12) bytes of memory?