Time period of oscillation of a bar magnet in a uniform magnetic field is \( T_o \). The magnet is cut into 3 equal parts transverse to its length. The new time period of oscillation of one part will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{T_o}{3} \)
A bar magnet of magnetic moment M and moment of inertia I oscillating in a field B has the time period
\( \displaystyle T = 2\pi\sqrt{\frac{I}{MB}} \)
Cutting the magnet transverse to its length into three equal pieces keeps the pole strength m of each piece the same, but the length becomes \( L/3 \) and the mass \( \mathrm{m_0}/3 \) (the mass of the whole bar is taken as \( \mathrm{m_0} \)).
Magnetic moment (pole strength \( \times \) length)
\( \displaystyle M' = m\cdot\frac{L}{3} = \frac{M}{3} \)
Moment of inertia about the axis through the centre, perpendicular to the length (the bar is thin, so \( I = \mathrm{m_0}L^{2}/12 \)):
\( \displaystyle I' = \frac{(\mathrm{m_0}/3)(L/3)^{2}}{12} = \frac{1}{27}\cdot\frac{\mathrm{m_0}L^{2}}{12} = \frac{I}{27} \)
New time period
\( \displaystyle T' = 2\pi\sqrt{\frac{I/27}{(M/3)B}} = 2\pi\sqrt{\frac{I}{9MB}} = \frac{1}{3}\cdot2\pi\sqrt{\frac{I}{MB}} = \frac{T_o}{3} \)
Note: NTA picked this question from an NCERT Exemplar problem and changed the figures. Do not ignore the Exemplar while practising numericals.
NCERT Exemplar:
Magnetism and Matter – …
