Question

Magnetic field of the earth is $H = 0.3g$. A magnet is vibrating five oscillations per minute then the depreciation required in the magnetic field of the earth of increase time period up to ten oscillations per minute is
A) 2.25g
B) 0.6g
C) 0.9g
D) 0.12g

Answer 1

When a magnet is placed in any external magnetic field, it states rotating in that magnetic field. So, it states vibrating. A time period is also associated with a magnetic field.
Time period of oscillation of a magnet is inversely proportional to the square root of the magnetic field.
Hence, by comparing we can easily find a solution to the above problem.

Formula used:
When a magnet (magnetic dipole) is placed in any external magnetic field, a torque is started acting on the dipole.
$\tau = - m \times B$
$\Rightarrow \tau = - mB\sin \theta$................(i)
If magnet is rotate small angular displacement do so $\sin \theta \simeq \theta$
And $\tau = I\alpha$................(ii)
Where $\alpha$ is the angular acceleration and $I$ is the moment of inertia.
From equation (i) and (ii), we get
$- mB\sin \theta = I\alpha$
Angular acceleration is the rate of change of angular velocity. So,
$I\dfrac{{{d^2}\theta }}{{d{t^2}}} = - mB\theta$
$I{\omega ^2} = - mB\theta$......................(iii)
By simple harmonic motion as the magnet is in simple harmonic motion.
$\tau = - {\omega ^2}\theta$................................(iv)
From equation (iii) and equation (iv)
${\omega ^2} = \dfrac{{mB}}{I} \\\ \Rightarrow \omega = \sqrt {\dfrac{{mB}}{I}} \\\$
Now, the time period will be-
$T = \dfrac{{2\pi }}{\omega }$
Putting the value of $\omega$ in the above equation, we get-
$T = 2\pi \sqrt {\dfrac{I}{{mB}}}$....................(v)

Complete step by step solution:
We have a magnetic field in which a magnet is vibrating at some time period and going to find the change in magnetic field when vibration of the magnet will increase. So,
${H_1} = 0.3g$
${T_1} = 5 oscillations/\min$
${T_2} = 10 oscillations/\min$
${H_2} = ?$
The time period of the magnet in external magnetic field is
$T = 2\pi \sqrt {\dfrac{I}{{mH}}}$
Where $I$ is the moment of inertia and $m$is the magnetic dipole moment. Both are constant for a magnet.
Now, we have-
${T_1} = 2\pi \sqrt {\dfrac{I}{{m{H_1}}}}$ (a)
And ${T_2} = 2\pi \sqrt {\dfrac{I}{{m{H_2}}}}$ (b)
Dividing the equations (a) and equation (b), we get-
$\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{2\pi \sqrt {\dfrac{I}{{m{H_1}}}} }}{{2\pi \sqrt {\dfrac{I}{{m{H_2}}}} }}$
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{{{H_2}}}{{{H_1}}}}$
Now, substituting the values, we have-
$\Rightarrow \dfrac{5}{{10}} = \sqrt {\dfrac{{{H_2}}}{{0.3g}}} \\\ \Rightarrow \dfrac{1}{2} = \sqrt {\dfrac{{{H_2}}}{{0.3g}}} \\\$
Or we can write-
$\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\sqrt {\dfrac{{{H_2}}}{{0.3g}}} } \right)^2} \\\ \Rightarrow \dfrac{1}{4} = \dfrac{{{H_2}}}{{0.3g}} \\\ \therefore {H_2} = 0.075g \\\$
Hence,
Depreciation in the magnetic field $= 0.3g - 0.075g$
Depreciation in the magnetic field $= 0.225g$

Therefore option (A) is correct.

Note: The moment of inertia of anybody which is in SHM is constant and the magnetic dipole moment depends upon the area and number of loops in the magnet which is constant for any magnet. One thing to remember is that when we are solving problems by comparing we don’t have to convert all the values in the MKS system.