Question 41:
easy9. A short bar magnet of magnetic moment $0.4$ J T$^{-1}$ is placed in a uniform magnetic field of $0.16$ T. The magnet is in stable equilibrium when the potential energy is: (2011 Mains)
For stable equilibrium, the angle between $M$ and $B$ is $0^{\circ}$.
Potential energy $U = -MB \cos 0^{\circ} = - (0.4)(0.16)(1) = -0.064$ J.
Question 42:
easy12. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It: (2012 Pre)
At the geomagnetic poles, the Earth's magnetic field is entirely vertical, meaning the horizontal component $B_H$ is zero. A compass free to rotate only horizontally experiences no directing torque and will stay in any position.
Question 43:
easy13. Due to earth’s magnetic field, the charged cosmic rays particles (1997)
At the equator, charged particles travel perpendicular to the Earth's magnetic field lines and experience maximum deflecting force. Thus, they need greater kinetic energy to penetrate the field and reach the equator compared to the poles.
Question 44:
easy14. A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2$ sec in earth’s horizontal magnetic field of $24$ microtesla. When a horizontal field of $18$ microtesla is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be: (2010 Pre)
Initial field $B_1 = 24$ $\mu$T, $T_1 = 2$ s. Net new field $B_2 = 24 - 18 = 6$ $\mu$T.
Since $T \propto \frac{1}{\sqrt{B}}$, we have $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$. Therefore, $T_2 = 2 \times 2 = 4$ s.
Question 45:
easy15. A bar magnet having a magnetic moment of $2 \times 10^4$ J T$^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times 10^{-4}$ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is: (2009)
Work done, $W = MB(\cos \theta_1 - \cos \theta_2) = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - \frac{1}{2}) = \frac{MB}{2}$.
$W = \frac{1}{2} \times (2 \times 10^4) \times (6 \times 10^{-4}) = 6$ J.
Question 46:
easy16. A bar magnet is oscillating in the Earth’s magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled? (2003)
Time period is given by $T = 2\pi \sqrt{\frac{I}{MB}}$. Moment of inertia $I$ is directly proportional to mass $m$. If mass is quadrupled, $I$ becomes $4I$.
New period $T' = 2\pi \sqrt{\frac{4I}{MB}} = 2T$. The restoring torque is still $-MB \sin \theta$, so motion remains S.H.M.
Question 47:
easy17. Two bar magnets having same geometry with magnetic moments $M$ and $2M$, are firstly placed in such a way that their similar poles are same side then its time period of oscillation is $T_1$. Now the polarity of one of the magnet is reversed then time period of oscillation is $T_2$, then: (2002)
$T = 2\pi \sqrt{\frac{I}{M_{net} B_H}}$. Initially $M_{net} = 2M + M = 3M$, so $T_1 \propto \frac{1}{\sqrt{3M}}$.
After reversing polarity, $M_{net} = 2M - M = M$, so $T_2 \propto \frac{1}{\sqrt{M}}$. Therefore, $T_1 < T_2$.
Question 48:
easy18. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is given by (1995)
$W_{90^{\circ}} = MB(1 - \cos 90^{\circ}) = MB(1 - 0) = MB$.
$W_{60^{\circ}} = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB$.
Since $W_{90^{\circ}} = n W_{60^{\circ}}$, we have $MB = n(0.5MB) \Rightarrow n = 2$.
Question 49:
easy20. A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from: (2018)
When the rod is pushed up, work is done against gravity. To maintain the steady magnetic field against this change, the current source (battery) must do work. Thus, the energy ultimately comes from the current source.
Question 50:
easy21. The magnetic susceptibility is negative for: (2016-I)
Magnetic susceptibility is negative for diamagnetic materials because they develop an induced magnetization in the direction opposite to the applied magnetic field. It is positive for paramagnetic and ferromagnetic materials.