\(\therefore \quad ILB = mg\)
\(\therefore \quad\left(\frac{ V }{ R }\right) LB = mg\)
\(\therefore \quad V =\frac{ mgR }{ LB }\)
\(=\frac{\left(1 \times 10^{-3}\,kg \right)\left(10\,m / s ^2\right)(10 \Omega)}{(0.1\,m )\left(10^3 \times 10^{-4}\,T \right)}=10\,V\)