The magnet of vibration magnetometer is heate | vedclass

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Question

(b) $T = 2\pi \sqrt {\frac{I}{{M{B_H}}}} $ $==>$ $T \propto \frac{1}{{\sqrt M }}$ $==>$ $\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}

Vedclass · Accepted Answer

Increases by $25\%$

Vedclass · Answer

(b) $T = 2\pi \sqrt {\frac{I}{{M{B_H}}}} $ $==>$ $T \propto \frac{1}{{\sqrt M }}$ $==>$ $\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}} $
If ${M_1} = 100$ then ${M_2} = (100 - 36) = 64$

So $\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{64}}{{100}}} = \frac{8}{{10}}$ $==>$ ${T_2} = \frac{{10}}{8}{T_1} = 1.25\,{T_1}$
So $%$ increase in time period $= 25\%$

The magnet of vibration magnetometer is heated so as to reduce its magnetic moment by $36\%$. By doing this the periodic time of the magnetometer will

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