Question 61:
easy32. Above Curie temperature: (2006)
The Curie temperature is the critical temperature point at which a material's intrinsic permanent magnetic moments change state. Heating a ferromagnetic substance above this point turns it into a paramagnetic substance.
Question 62:
easy33. According to Curie’s law, the magnetic susceptibility of a substance at an absolute temperature $T$ is proportional to: (2003)
Curie's law states that the magnetic susceptibility ($\chi$) of a paramagnetic substance is inversely proportional to its absolute temperature ($T$). Therefore, $\chi \propto 1/T$.
Question 63:
easy5. A closely wound solenoid of $2000$ turns and area of cross section $1.5 \times 10^{-4} m^2$ carries a current of $2.0 A$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2} tesla$ making an angle of $30^{\circ}$ with the axis of the solenoid. The torque on the solenoid will be (2010 Mains)
Magnetic moment of the solenoid is $M = N I A = 2000 \times 2.0 \times (1.5 \times 10^{-4}) = 0.6 J/T$. The torque acting on the solenoid is $\tau = M B \sin\theta = 0.6 \times (5 \times 10^{-2}) \times \sin(30^{\circ}) = 1.5 \times 10^{-2} Nm$.
Question 64:
easy4. A bar magnet of magnetic moment $M$ is cut into two parts of equal length. The magnetic moment of each part will be (1997)
When a magnet of magnetic moment $M = m \times l$ is cut into two parts of equal length, the pole strength $m$ remains the same while the length becomes $l/2$. Thus, the new magnetic moment of each part is $M' = m \times (l/2) = M/2 = 0.5 M$.
Question 65:
easy7. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{circ}$ is $W$. Now the torque required to keep the magnet in this new position is: (2016 – II)
Work done in rotating the magnet from equilibrium is $W = MB(1 - cos 60^{circ}) = frac{MB}{2}$, giving $MB = 2W$. The torque required in this position is $tau = MB sin 60^{circ} = (2W)left(frac{sqrt{3}}{2}right) = sqrt{3}W$.
Question 66:
easy8. A magnetic needle suspended parallel to a magnetic field requires $sqrt{3} text{ J}$ of work to turn it through $60^{circ}$. The torque needed to maintain the needle in this position will be: (2012 Mains)
Work done is $W = MB(1 - cos 60^{circ}) = frac{MB}{2} = sqrt{3}$, which implies $MB = 2sqrt{3} text{ J}$. The torque required is $tau = MB sin 60^{circ} = 2sqrt{3} times frac{sqrt{3}}{2} = 3 text{ J}$.
Question 67:
easy9. A short bar magnet of magnetic moment $0.4 text{ J T}^{-1}$ is placed in a uniform magnetic field of $0.16 text{ T}$. The magnet is in stable equilibrium when the potential energy is: (2011 Mains)
For stable equilibrium, the angle between $vec{M}$ and $vec{B}$ is $theta = 0^{circ}$. The potential energy is $U = -MB cos 0^{circ} = -0.4 times 0.16 = -0.064 text{ J}$.
Question 68:
easy13. Due to earth’s magnetic field, the charged cosmic rays particles (1997)
At the equator, magnetic field lines are perpendicular to incoming charged particles, causing maximum magnetic deflection force. At the poles, field lines are nearly parallel, causing little to no deflection. Thus, particles require much higher energy to penetrate at the equator.
Question 69:
easy14. A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2text{ sec}$ in earth’s horizontal magnetic field of $24text{ microtesla}$. When a horizontal field of $18text{ microtesla}$ is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be: (2010 Pre)
Time period is $T propto frac{1}{sqrt{B}}$. Initially $B_1 = 24 mutext{T}$. In the second case, the net field is $B_2 = 24 - 18 = 6 mutext{T}$. Therefore, $T_2 = T_1 sqrt{frac{B_1}{B_2}} = 2 times sqrt{frac{24}{6}} = 2 times 2 = 4text{ s}$.
Question 70:
easy15. A bar magnet having a magnetic moment of $2 times 10^4text{ JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 times 10^{-4}text{ T}$ 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 is given by $W = MB(1 - costheta)$. Substituting the values: $W = (2 times 10^4) times (6 times 10^{-4}) times (1 - cos 60^{circ}) = 12 times 0.5 = 6text{ J}$.