TS EAMCET 14-Sep-2020 Shift 1 Solved Paper

∵τ=MB‌sin‌θ
where, τ= torque experienced by bar magnet
=0.016N−m
M= magnetic moment of bar magnet
B= external magnetic field intensity
θ= angle made by the axis of bar magnet with the external magnetic field =30∘
So, 0.016=M×8×10−2×sin(30∘)
⇒0.016=M×8×10−2×‌

1

2

⇒M=0.4A−m2