Question 261:
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 $\mu_{f} \neq 0$.
Question 262:
easy26. Diamagnetic material in a magnetic field moves: (2003)
Diamagnetic materials get magnetized in the opposite direction to the applied external magnetic field. Because of this repulsion, they experience a net force moving them from the stronger region to the weaker region of the field.
Question 263:
easy27. Among which the magnetic susceptibility does not depend on the temperature: (2001)
According to Curie's law, the susceptibility of paramagnetic and ferromagnetic materials varies inversely with temperature. However, the magnetic susceptibility of diamagnetic materials is practically independent of temperature.
Question 264:
easy28. For protecting a magnetic needle it should be placed: (1998)
Soft iron is a ferromagnetic material with high permeability. When an iron box is placed in an external magnetic field, most of the magnetic field lines pass through the walls of the box, shielding the inside space.
Question 265:
easy29. Electromagnets are made of soft iron because soft iron has: (2010 Pre)
Electromagnets require a material that gets magnetized easily and quickly loses its magnetism when the current is turned off. Soft iron is ideal because it has low retentivity and low coercive force.
Question 266:
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 267:
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 268:
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 269:
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 270:
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 $U = -MB cos 0^{circ} = -0.4 times 0.16 = -0.064 text{ J}$.