A cylindrical bar magnet whose mass is 0.03 kg, diameter is 1 cm, length is 3 cm

Question

A cylindrical bar magnet whose mass is 0.03 kg, diameter is 1 cm, length is 3 cm, and whose magnetic dipole moment is (8, 0, 0) A m^2 is suspended on a low-friction pivot in a region where external coils apply a magnetic field of (1.4, 0, 0) T. You rotate the bar magnet slightly in the horizontal plane and release it. (For small angles in radians, assume sin(theta) theta.) What is the angular frequency of the oscillating magnet? rad/s What would be the angular frequency if the applied magnetic field were (2.8, 0, 0) T? rad/s

Explanation / Answer

moment of inertia is,

   I = 0.25mR2 + (1/12)mL2

     = 0.25*0.03*[0.5x10-2]2 + (1/12)*0.03*[0.03]2

    = 2.4375e-6 kg.m2

--------------------------------------------------------------------------------

angular frequency is,

   w = sqrt[MB/I] = sqrt[8*1.4 / 2.4375e-6] = 2143.56 rad/s

---------------------------------------------------------------------------------

angular frequency:

   w = sqrt[MB/I] = sqrt[8*2.8 / 2.4375e-6] = 3031.46 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.