JEE Main20204 Sep 2020Morning ShiftPhysicsMagnetic Properties of MatterActual
A small bar magnet is placed with its axis at 30 o with an external magnetic field of 0 . 06 T experiences a torque of 0 . 018 Nm . The minimum work required to rotate it from its stable to unstable equilibrium position is:
Options
- A6 . 4 × 10 − 2   J
- B9 . 2 × 10 − 3   J
- C7 . 2 × 10 − 2   J
- D11 . 7 × 10 − 3   J
Correct answer
C. 7 . 2 × 10 − 2   J
Step-by-step solution
τ = MBsinθ = 0 . 018 M = 0 .018 B   sin   θ = 0 .018 0 .06 × 0 .5 = 0 .64   A m 2 W = ΔU = U f − U i = − MB   cos   180 ∘ − − MB   cos 0 ∘ = 2 MB = 2 × 0 .6 × 0 .06 = 0 . 072   J