Ncert Exercises & Solutions

A bar magnet of magnetic moment M and moment of inertia I (about the centre, perpendicular to length) is cut into two equal pieces, perpendicular to the length. Let T be the period of oscillations of the original magnet about an axis through the mid-point, perpendicular to the length, in a magnetic field B. What would be a similar period T' for each piece?

Hint: The time period depends on the magnetic moment and the moment of inertia of the magnet.

Given, I= moment of inertia of the bar magnet about an axis passing through its centre and perpendicular to its length
m= mass of bar magnet
l= length of the magnet
M =magnetic moment of the magnet
B = uniform magnetic field in which the magnet is oscillating

Step 1: The time period of oscillation is:

T = 2 π \sqrt{\frac{I}{MB}}

Here, I = \frac{{ml}^{2}}{12}

Step 2: When the magnet is cut into two equal pieces, perpendicular to the length, then the moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is:

I^{'} = \frac{m}{2} \frac{(l / 2)^{2}}{12} = \frac{{ml}^{2}}{12} \times \frac{1}{8} = \frac{I}{8}

Magnetic dipole moment M' = M/2

Step 3: Its time period of oscillation is:

\begin{array}{l} T^{'} = 2 π \sqrt{\frac{I^{'}}{M^{'} B}} = 2 π \sqrt{\frac{I / 8}{(M / 2) B}} = \frac{2 π}{2} \sqrt{\frac{I}{MB}} \\ T^{'} = \frac{T}{2} \end{array}

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