(D) The time period $T$ of oscillation of a magnet is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic m

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(D) The time period $T$ of oscillation of a magnet is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the horizontal magnetic field.Since $I$ and $M$ are constant,we have $T \propto \frac{1}{\sqrt{B}}$,which implies $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$.Given $B_1 = 24 \, \mu T$ and $T_1 = 2 \, s$.The new magnetic field is $B_2 = B_1 - 18 \, \mu T = 24 \, \mu T - 18 \, \mu T = 6 \, \mu T$.Substituting the values: $\frac{T_2}{2} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$.Therefore,$T_2 = 2 \times 2 = 4 \, s$.

Class 12 Physics Magnetism and Matter Magnetic Equipment's (Tangent galvanometer, Vibration magnetometer and Neutral point)

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$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$

AIPMT 2010 All AIPMT

A
$1$
B
$2$
C
$3$
D
$4$

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Class 12 Questions

Class 12

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Magnetism and Matter

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Magnetic Equipment's (Tangent galvanometer, Vibration magnetometer and Neutral point)

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$A$ magnet makes $40$ oscillations per minute at a place having a magnetic intensity of $0.1 \times 10^{-5} \, T$. At another place, it takes $2.5 \, s$ to complete one oscillation. The value of the Earth's horizontal magnetic field at that place is:

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$A$ 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 $, s$)

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Two magnets are placed with their like poles together,and they perform $12$ oscillations in $1$ minute. When they are placed with their unlike poles together,they perform $4$ oscillations in $1$ minute. What is the ratio of the magnetic moments of the magnets?

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In the sum and difference method of a vibration magnetometer,the time period is greater if:

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$A$ bar magnet is oscillating in the earth's magnetic field with a time period $T$. If a similar magnet with the same mass and volume has a magnetic dipole moment which is $9$ times that of this magnet,then its time period will be . . . . . . .

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$A$ magnet makes $40$ oscillations per minute at a place having a magnetic intensity of $0.1 \times 10^{-5} \, T$. At another place, it takes $2.5 \, s$ to complete one oscillation. The value of the Earth's horizontal magnetic field at that place 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 $, s$)

Two magnets are placed with their like poles together,and they perform $12$ oscillations in $1$ minute. When they are placed with their unlike poles together,they perform $4$ oscillations in $1$ minute. What is the ratio of the magnetic moments of the magnets?

In the sum and difference method of a vibration magnetometer,the time period is greater if:

$A$ bar magnet is oscillating in the earth's magnetic field with a time period $T$. If a similar magnet with the same mass and volume has a magnetic dipole moment which is $9$ times that of this magnet,then its time period will be . . . . . . .

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