The magnetic needle of a vibration magnetometer makes 12...

The magnetic needle of a vibration magnetometer makes 12 oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line, it makes 15 oscillations per minute. If the poles of the bar magnet are interchanged, the number of oscillations it makes per minute is

\sqrt{n}

. What is the value of

n

Correct answer

Step-by-step solution

Ist case :

n = \dfrac{1}{2\pi}\sqrt{\dfrac{M{B}_{H}}{I}}

⇒ n ∝ \sqrt{{B}_{H}} ⇒ \dfrac{{n}_{1}}{{n}_{2}} = \sqrt{\dfrac{{B}_{H}}{{B}_{H} + {{B}_{H}}_{1}}}

⇒ \dfrac{12}{15} = \sqrt{\dfrac{{B}_{H}}{{B}_{H} + {{B}_{H}}_{1}}}

⇒ {{B}_{H}}_{1} = \dfrac{9}{16}{B}_{H}

IInd case :

\dfrac{{n}_{2}}{{n}_{3}} = \sqrt{\dfrac{{B}_{H} + {{B}_{H}}_{1}}{{B}_{H} - {{B}_{H}}_{1}}}

⇒ \dfrac{15}{{n}_{3}} = \sqrt{\dfrac{{B}_{H} + \dfrac{9}{16}{B}_{H}}{{B}_{H} - \dfrac{9}{16}{B}_{H}}}

∴ {n}_{3} = \sqrt{63} ⇒ n = 63

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