$I=q f.........(i)$
The magnetic induction at the centre of the ring is
$B=\frac{\mu_{0} 2 \pi I}{4 \pi R}=\frac{\mu_{0} q f}{2 R} \quad(\text { Using }(\mathrm{i}))$


Statement $-2$ : The magnetic field due to finite length of a straight current carrying wire is symmetric about the wire.