Here is a construction of size $38$. Call the points cards and divide the deck into $4$ suits, so that $2$ suits have ranks $1,...,13$ and $2$ suits have ranks $1,...,12$. Every hand of $5$ cards must have at least $2$ cards in some suit. So, if we cover all possible pairs in each suit, the result will be a wheel.
The La Jolla Covering Repository says that $C(12,5,2)=9$ so we can cover all pairs within the short suits with $9$ hands each:
1 2 4 7 9
1 3 4 6 12
1 3 7 10 11
1 5 6 8 12
2 3 5 7 8
2 6 10 11 12
3 6 7 9 12
4 5 8 10 11
5 8 9 10 11
Similarly, it says $C(13,5,2) \le 10$ so we can cover all pairs within the long suits with $10$ hands each:
1 2 3 4 5
1 6 7 8 9
1 10 11 12 13
2 3 6 7 10
2 3 8 9 11
4 5 6 7 11
4 5 8 9 10
2 3 4 12 13
5 6 7 12 13
1 8 9 12 13
The total is $38$ hands which intersect each hand of $5$ cards in at least $2$ cards. So, $L(50,5,5,2) \le 38$.
Edit: The best this method can do with the coverings known in the La Jolla Repository is $37$. That can be done by splitting a $50$ card deck into suits of sizes $(17, 11, 11, 11),$ $(13,13,13,11)$, or $(15, 13, 11, 11)$.
$C(13,5,2) \le 10$ as shown above.