The magnetic intensity H inside the solenoid is,
H=ni
⇒H=2000×2=4000 Am−1
We know that,
B=μ0(H+I)
⇒I=Bμ0−H=1.574π×10−7−4000
=1.25×106−4000
≈1.25×106 Am−1
The pole strength developed at the end is,
m=IA
=(1.25×106)×(5×10−4)
=625 Am
Hence, option (D) is the correct answer.