A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19 %. The periodic time of the magnetometer will

Question

A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19 %. The periodic time of the magnetometer will (A) increase by 19 % (B) decrease by 19 % (C) increase by 11 % (D) decrease by 11 %

Tardigrade Admin · Accepted Answer

T=2 π √(I/M B) and T'=2 π √(I × 100/81 M B) ⇒ T'=(10/9) T Percentage increase = ((10/9)-1/1) × 100 =(1/9) × 100 ≈ 11 %

Q. A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by $19%$ . The periodic time of the magnetometer will

increase by $19%$

decrease by $19%$

increase by $11%$

decrease by $11%$

$T = 2 π \frac{I}{MB}$
and $T^{'} = 2 π \frac{I \times 100}{81 MB}$
$\Rightarrow T^{'} = \frac{10}{9} T$
Percentage increase $= \frac{\frac{10}{9} - 1}{1} \times 100$
$= \frac{1}{9} \times 100 \approx 11%$

Q. A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. The periodic time of the magnetometer will

increase by 19%

decrease by 19%

increase by 11%

decrease by 11%