Two bar magnets $A$ and $B$ are placed one over the other and are allowed to vibrate in a vibration…

Two bar magnets $A$ and $B$ are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make 20 oscillations per minute when the similar poles of $A$ and $B$ are on the same side, while they make 15 oscillations per minute when their opposite poles lie on the same side. If $M_A$ and $M_B$ are the magnetic moments of $A$ and $B$ and if $M_A>M_B$, the ratio of $M_A$ and $M_B$ is
  1. $4: 3$
  2. $25:7$
  3. $7:5$
  4. $25:16$

Solution

Ratio of magnetic moments of two magnets of equal size when in sum and difference position is $\frac{M_A}{M_B}=\frac{T_d^2+T_s^2}{T_d^2-T_s^2}=\frac{v_s^2+v_d^2}{v_s^2-v_d^2}$ $\begin{aligned} & =\frac{\left(\frac{1}{20}\right)^2+\left(\frac{1}{15}\right)^2}{\left(\frac{1}{15}\right)^2-\left(\frac{1}{20}\right)^2} \\ & =\frac{400+225}{400-225} \\ & =\frac{625}{175}=\frac{25}{7}\end{aligned}$ $\Rightarrow \quad M_A: M_B=25: 7$

Asked in: AP EAMCET 2009

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