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Practice each question, then open the answer or worked solution.
A proton has a spin and magnetic moment, just like an electron. Then, why is its effect neglected in the magnetism of materials?
The comparison between the spinning of a proton and an electron is done by comparing their magnetic dipole moment, which is given as
μp = eh/4πmp
μe = eh/4πme
μp/μe = me/mp = 1/1837 >> 1
μp << μe
A permanent magnet in the shape of a thin cylinder of length 10 cm has M = 106 A/m. Calculate the magnetisation current I m.
Intensity of magnetisation = 106 A/m
Length, l = 0.1 m
M = IM/l
IM = Ml = 105 A
Explain the order of magnitude difference between the diamagnetic susceptibility of N2 and Cu quantitatively.
The density of nitrogen = 28 g/ 22400 cc
The density of copper = 8 g/ 22400 cc
Ratio of densities = 16 × 10-4
Diamagnetic susceptibility = density of nitrogen/density of copper = 1.6 × 10-4
From a molecular viewpoint, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism, and ferromagnetism.
The temperature dependence of susceptibility for a diamagnetism is not much affected by the temperature.
The temperature dependence of susceptibility for paramagnetism and ferromagnetism is affected by the temperature that is, as the temperature rises, the magnetic moments get disturbed.
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. i) In which direction will it move? ii) What will be the direction of its magnetic moment?
i) The superconducting material will move away from the bar magnet.
ii) The direction of the magnetic moment will be from left to right.
Short AnswersVerify Gauss’s law for the magnetic field of a point dipole of dipole moment m at the origin for the surface, which is a sphere of radius R.
P is the point at a distance r from O and OP, then the magnetic field is given as:
dS is the elementary area of the surface P, then
dS = r2 (r2 sin θ d θr)
Solving the above, we get,
Three identical bar magnets are riveted together at the centre in the same plane, as shown in the figure. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet are shown in the figure. Determine the poles of the remaining two.
The system will have a net torque and the net force equal to zero as the system is in equilibrium.
Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of i) electric dipole p in an electrostatic field E and ii) magnetic dipole m in a magnetic field B. Write down a set of conditions on E, B, p, m so that the two movements are verified to be identical.
pE sin θ = μB sin θ
pE = μB
E = cB
pcB = μB
p = μ/c
A bar magnet of magnetic moment m and moment of inertia I are cut into two equal pieces, perpendicular to the length. Let T be the period of oscillations of the original magnet about an axis through the midpoint, perpendicular to the length, in a magnetic field B. What would be the similar period T’ for each piece?
T is the time period
I is the moment of inertia
m is the mass of the magnet
B is the magnetic field
T = 2π√I/MB
Magnetic dipole moment M’ = M/2
The time period is given as T’ = T/2
Use i) the Ampere’s law for H and ii) the continuity of lines of B to conclude that inside a bar magnet, a) lines of H run from the N pole to the S pole, while b) lines of B must run from the S pole to N pole.
C is the Amperian loop which is given as
Solving the above equation, we get the angle between
more than 90o so that cos θ is negative.