STANDARD Physics Magnetic Dipole MCQ Question
The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole = 0.5 A m²)
0.07 J
0.08 J
0.09 J
0.1 J
Correct Answer
Detailed Explanation
Since the most stable position is at θ = 0 and the most unstable position is at θ = 180°, the work done is given by W = -MB[cos 180° - cos 0°] = -MB[-1 - 1] = 2MB. Here, M = 0.5 A m² and B = 0.09 T. Therefore, W = 2 × 0.5 × 0.09 = 0.09 J.
Found an issue with this question?
Related Questions
More from Magnetism
Assertion: Paramagnetic substances get poorly attracted in magnetic field.
The magnetic force on the charged particle is given by, F_B = qvBsinθ. As the particle is released from rest therefore, there may be no force on the p...
At what distance for a long straight wire carrying a current of 12 Å will the magnetic field be equal to 3×10⁻⁵ Wb/m²?