Given:
R=0.01 Ω; N=400; B=10−2 T
r1=2 cm; r2=3 cm
rmean=r1+r22=2.5 cm=2.5×10−2 m
Let the emf of the battery be E.
Current flowing through the coil is,
I=ER=E0.01=100E
The magnetic field inside the toroid is,
B=μ0NI2πr
10−2=4π×10−7×400×100E2π×2.5×10−2
E=2.5×10−2×10−22×10−7×4×104=3.125×10−2 V
∴E=31.25 mV
Hence, option (A) is the correct answer.