
Two magnets of same size and mass make, respectively, 10 and 15 oscillations per minute at a certain place. The ratio of their magnetic moments is:
A. $4:9$
B. $9:4$
C. $2:3$
D. $3:2$
Answer
162.9k+ views
Hint: A magnet's magnetic moment interacts with an applied field to produce a mechanical moment. Remember the concepts of torque and magnetic field. A vector quantity is a magnetic moment. We will outline the relationship between frequency and magnetic moment in order to answer this question.
Formula used:
The expression of frequency ($v$) of magnet is,
$v = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{mB}}{I}}$
Here, $m$ is the magnetic moment, $B$ is the magnetic field and $I$ is the moment of inertia.
Complete step by step solution:
A magnet's magnetic moment interacts with an applied field to produce a mechanical moment. Remember the concepts of torque and magnetic field. A vector quantity is a magnetic moment. We will outline the relationship between frequency and magnetic moment in order to answer this question.
The motion of the electric charge method and the spin angular momentum method can both be used to produce the magnetic moment. A device known as a magnetometer is often used to measure an object's magnetic moment.
In this question the magnetic field component will be the same and frequency is inverse of the time period of the magnet. The frequency of magnet is given by,
$v = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{mB}}{I}} $
When we increase the frequency, the magnetic moment increases as the square of frequency.
Given that no. of oscillation by ${m_1} = 10$ per minute
No. of oscillation by ${m_2} = 15$ per minute
So we can say that $v \propto \sqrt m $
$ \Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $
$ \Rightarrow \dfrac{{10}}{{15}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $
$ \therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{4}{9}$
Hence option A is correct.
Note: The highest torque a magnet will experience in a unit external magnetic field is referred to as the magnetic moment's magnitude. Due to some torque the magnet experiences when it is placed in the external magnetic field, which is caused by the magnetic dipole, the magnet strives to align itself with the external magnetic field.
Formula used:
The expression of frequency ($v$) of magnet is,
$v = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{mB}}{I}}$
Here, $m$ is the magnetic moment, $B$ is the magnetic field and $I$ is the moment of inertia.
Complete step by step solution:
A magnet's magnetic moment interacts with an applied field to produce a mechanical moment. Remember the concepts of torque and magnetic field. A vector quantity is a magnetic moment. We will outline the relationship between frequency and magnetic moment in order to answer this question.
The motion of the electric charge method and the spin angular momentum method can both be used to produce the magnetic moment. A device known as a magnetometer is often used to measure an object's magnetic moment.
In this question the magnetic field component will be the same and frequency is inverse of the time period of the magnet. The frequency of magnet is given by,
$v = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{mB}}{I}} $
When we increase the frequency, the magnetic moment increases as the square of frequency.
Given that no. of oscillation by ${m_1} = 10$ per minute
No. of oscillation by ${m_2} = 15$ per minute
So we can say that $v \propto \sqrt m $
$ \Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $
$ \Rightarrow \dfrac{{10}}{{15}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $
$ \therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{4}{9}$
Hence option A is correct.
Note: The highest torque a magnet will experience in a unit external magnetic field is referred to as the magnetic moment's magnitude. Due to some torque the magnet experiences when it is placed in the external magnetic field, which is caused by the magnetic dipole, the magnet strives to align itself with the external magnetic field.
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