Question 141:
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 142:
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 143:
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 144:
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 145:
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 146:
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 147:
easy19. An iron rod of susceptibility $599$ is subjected to a magnetising field of $1200$ A m$^{-1}$. The permeability of the material of the rod is : (2020)
($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)
Relative permeability $\mu_r = 1 + \chi = 1 + 599 = 600$.
Permeability $\mu = \mu_r \mu_0 = 600 \times 4\pi \times 10^{-7} = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4}$ T m A$^{-1}$.
Question 148:
easy23. The magnetic moment of a diamagnetic atom is: (2010 Mains)
In diamagnetic materials, electrons occur in pairs with opposite spins and their orbital motions cancel each other out. Thus, their individual magnetic moments cancel, resulting in a net magnetic moment equal to zero for the atom.
Question 149:
easy24. If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is: (2009)
Diamagnetic substances always move from stronger parts to weaker parts of a non-uniform magnetic field. Therefore, they are repelled by both the north and south poles of a bar magnet.
Question 150:
easy25. If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are denoted by $\mu_{d}$, $\mu_{p}$ and $\mu_{f}$ respectively, then (2005)
Diamagnetic atoms have completely paired electrons, meaning no permanent magnetic dipole moment, so $\mu_{d} = 0$. Paramagnetic and ferromagnetic atoms have unpaired electrons and possess permanent dipole moments, so $\mu_{p} \neq 0$ and $\mu_{f} \neq 0$.