Why a light metal disc on the top of an electromagnet is thrown up as the current is switched ON. ?

Why a light metal disc on the top of an electromagnet is thrown up as the current is switched ON. ?

Sol.
When the current begins to grow through the electromagnet, the magnetic flux through the disc begins to increase. This sets up eddy current in the disc in the same direction as that of the electromagnetic current.
Thus, if the upper surface of electromagnet acquires N- polarity, the lower surface of the disc also acquires N-polarity. As, same magnetic poles repel each other, the light metallic disc is thrown up.

Why is the coil of dead beat galvanometer wound on a metal frame ?

Q: Why is the coil of dead beat galvanometer wound on a metal frame ?

Sol.
On switching ON the current in a galvanometer, the coil of the galvanometer does not come to rest immediately. It oscillates about its equilibrium position but the coil of a dead bear galvanometer comes to rest immediately. It is due to the reason that the eddy currents are set up in the metallic frame,over which the coil is would and the eddy current oppose the oscillatory motion of the coil.

Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled.?

Answer the following questions :

(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled.

(b) Why is diamagnetism, in contrast, almost independent of temperature?

(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty ?

(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

(e) Magnetic field lines are always normal to the surface of a ferromagnet at every point (this fact is analogous to the static electric lines being normal to the surface of a conductor at every point) why ?

(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet ?

Sol.
(a) This is because at lower temperatures, the tendency to disrupt the alignment of dipoles (due to magnetising field) decreases on account of reduced random thermal motion.

(b) In a diamagnetic sample, each molecule is not a magnetic dipole in itself. Therefore, random thermal motion of molecules does not affect the magnetism of the specimen. That is why diamagnetism is independent of temperature.

(c) As bismuth is diamagnetic, therefore, the field in the core in the core will be slightly less than when the core is empty.

(d) No. permeability of a ferromagnetic material is not independent of magnetic field. As is clear from the hysteresis curve, μ is greater for lower fields.

(e) Magnetic field lines area always nearly normal to the surface of a ferromagnet at every point. The proof of this important fact is based on the boundary conditions of magnetic fields (B and H) at the interface of two media. The magnetic permeability of a ferromagnetic material μ > > 1. That is why the field lines meet this medium normally.

(f) Yes. Maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet. The saturation however, requires very high magnetising fields, which are hard to achieve.

Why speed of sound wave in air depends on temperature but that of light does not ?

Sol: Sound is a mechanical wave and for which

$\displaystyle V = \sqrt{\frac{\gamma R T}{m}}$

while light is a nonmechanical wave for which it

$ \displaystyle C = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$

This last but very important point is that – any function of space and time. [y(x,t)] which satisfies the equation-

$ \displaystyle \frac{d^2 y}{dx^2} = \frac{1}{V^2} \frac{d^2 y}{d t^2}$

Represents a wave, with speed V e.g. functions Y = A sin ωt or y = A Sin kx do not satisfy the above equation so do not represent waves, while functions A log (ax+bt), A Sin(ωt-kx), A sin kx Sin ωt or A sin (ωt-kx)+ B Cos (ωt+kx) satisfy the above equation so represent waves.

It can also be shown that the equation of a travelling wave is of the form:

Y = f (at – bx)

Negative sign between at and bx implies that the wave is travelling along position x-axis and vice-versa
If a travelling wave is a sin or cos function of (at – bx) or (at+bx), The wave is said to be harmonic or plane progressive wave

It can also be shown for a plane progressive wave is-

y = A sin ( ωt-kx)

$\displaystyle \frac{dy}{dt} = – \frac{\omega}{k} \frac{dy}{dx} $

$ \displaystyle \frac{dy}{dt} = – \frac{2\pi \nu}{2\pi/\lambda} \frac{dy}{dx} $

$\displaystyle \frac{dy}{dt} = – \nu \lambda \frac{dy}{dx} $

$ \displaystyle \frac{dy}{dt} = – V \frac{dy}{dx} $

i.e. Particle velocity = – (wave speed) (slope of wave at that point)

Vp = – V (slope)

Why is the shape of a rotating liquid surface Paraboloidal ?

Q: Why is the shape of a rotating liquid surface Paraboloidal ?

Ans: In a rotating liquid there is a pressure gradation from the axis to the wall . The pressure at a distance x from the axis is given by

$ \displaystyle p = p_0 + \frac{1}{2} \rho \omega^2 x^2 $

Consider a point P at a distance x from the axis and at a depth y from the liquid surface . Then

$\displaystyle p = p_0 + \rho g y $

Hence , $ \displaystyle p_0 + \rho g y = p_0 + \frac{1}{2} \rho \omega^2 x^2 $

$ \displaystyle y = \frac{\omega^2}{2 g} x^2 $

or , $ \displaystyle x^2 = ( \frac{2 g}{\omega^2} ) y $

On comparing this equation with the standard form of equation of a parabola x2 = 4 a y , we find that rotating liquid surface is a parabola about the vertical axis of rotation and the entire surface is therefore , Paraboloidal .