(C) The frequency of oscillation $n$ for a magnet in a horizontal magnetic field $B_H$ is given by the formula: $n = \frac{1}{2\pi} \sqrt{\frac{MB_H}{

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(C) The frequency of oscillation $n$ for a magnet in a horizontal magnetic field $B_H$ is given by the formula: $n = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$,where $M$ is the magnetic moment and $I$ is the moment of inertia.From this relation,we can see that $n \propto \sqrt{MB_H}$.Given that the magnetic moment becomes $M' = 4M$ and the magnetic field becomes $B_H' = 2B_H$.Substituting these values into the proportionality,we get the new frequency $n'$:$n' \propto \sqrt{M' B_H'} = \sqrt{(4M)(2B_H)} = \sqrt{8MB_H} = 2\sqrt{2} \sqrt{MB_H}$.Since $n \propto \sqrt{MB_H}$,it follows that $n' = 2\sqrt{2} n$.

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

Medium

$A$ magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled,the number of oscillations made per minute would be

A
$\frac{n}{2\sqrt{2}}$
B
$\frac{n}{\sqrt{2}}$
C
$2\sqrt{2} n$
D
$\sqrt{2} n$

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Magnetism and Matter

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

$A$ 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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$A$ long vertical wire in which a current is flowing produces a neutral point with the Earth's magnetic field at a distance of $5 \ cm$ from the wire. If the horizontal component of the Earth's magnetic induction is $0.18 \ G$,then the current in the wire is: (in $A$)

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$A$ current-carrying coil is placed with its axis perpendicular to the $N-S$ direction. Let the horizontal component of the Earth's magnetic field be $H_0$ and the magnetic field inside the loop be $H$. If a magnet is suspended inside the loop,it makes an angle $\theta$ with $H$. Then $\theta =$

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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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At a certain place,a magnet makes $30$ oscillations per minute. At another place where the magnetic field is doubled,its time period will be

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$A$ magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled,the number of oscillations made per minute would be

Similar Questions

$A$ 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 . . . . . . .

$A$ long vertical wire in which a current is flowing produces a neutral point with the Earth's magnetic field at a distance of $5 \ cm$ from the wire. If the horizontal component of the Earth's magnetic induction is $0.18 \ G$,then the current in the wire is: (in $A$)

$A$ current-carrying coil is placed with its axis perpendicular to the $N-S$ direction. Let the horizontal component of the Earth's magnetic field be $H_0$ and the magnetic field inside the loop be $H$. If a magnet is suspended inside the loop,it makes an angle $\theta$ with $H$. Then $\theta =$

$A$ 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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