A vibration magnetometer consists of two iden | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic mo

Vedclass · Accepted Answer

$2^{7/4}$

Vedclass · Answer

(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal magnetic field.For the combination of two identical magnets (each with moment of inertia $I_0$ and magnetic moment $M_0$) placed perpendicularly and bisecting each other:Total moment of inertia $I_1 = I_0 + I_0 = 2I_0$.Resultant magnetic moment $M_1 = \sqrt{M_0^2 + M_0^2} = M_0\sqrt{2}$.Given $T_1 = 4 \, s$,so $4 = 2\pi \sqrt{\frac{2I_0}{M_0\sqrt{2} B_H}} = 2\pi \sqrt{\frac{I_0\sqrt{2}}{M_0 B_H}}$.When one magnet is removed:New moment of inertia $I_2 = I_0$.New magnetic moment $M_2 = M_0$.New time period $T_2 = 2\pi \sqrt{\frac{I_0}{M_0 B_H}}$.Taking the ratio:$\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{I_0\sqrt{2}}{M_0 B_H}}}{2\pi \sqrt{\frac{I_0}{M_0 B_H}}} = \sqrt{\sqrt{2}} = 2^{1/4}$.Therefore,$T_2 = \frac{T_1}{2^{1/4}} = \frac{4}{2^{1/4}} = \frac{2^2}{2^{1/4}} = 2^{2 - 1/4} = 2^{7/4} \, s$.

Similar Questions

Time period of a freely suspended magnet does not depend upon

When $2 \ A$ current is passed through a tangent galvanometer,it gives a deflection of $30^\circ$. For $60^\circ$ deflection,the current must be........$A$.

$A$ vibration magnetometer is used at two different places $A$ and $B$ on the earth. The time period of a magnet suspended freely in the magnetometer at $A$ is twice that at $B$. If the horizontal component of the earth's magnetic field at $B$ is $32 \times 10^{-6} \,T$, then its value at $A$ is

$A$ short magnet oscillates with a time period $0.1 \, s$ at a place where the horizontal magnetic field is $24 \, \mu T$. $A$ downward current of $18 \, A$ is established in a vertical wire $20 \, cm$ east of the magnet. Find the new time period of the oscillator. (in $, s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module