A small bar magnet placed with its axis at 30° with an external field of 0.06 T experiences a torque of 0.018Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:
A small bar magnet placed with its axis at 30° with an external field of 0.06 T experiences a torque of 0.018Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:
Option 1 -
11.7 * 10⁻³ J
Option 2 -
9.2 * 10⁻³ J
Option 3 -
7.2 * 10⁻² J
Option 4 -
6.4 * 10⁻² J
Asked by Shiksha User
-
1 Answer
-
Correct Option - 4
Detailed Solution:τ? = μ? × B?
⇒ 0.018 = μ(0.06)(sin 30°)
⇒ μ = 0.6
⇒ Work = Uf – Ui
⇒ 2μB
⇒ 7.2 × 10?² J.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers