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The simplest magnetic element known to exist is a/an:
The NCERT text states: 'There are no sources or sinks of B; the simplest magnetic element is a dipole or a current loop.'
Why is the net magnetic flux through any closed surface zero?
The text explains this: 'The difference between the Gauss’s law of magnetism and that for electrostatics is a reflection of the fact that isolated magnetic poles (also called monopoles) are not known to exist.' Since magnetic field lines don't begin or end at points (like electric field lines do on charges), they must form continuous closed loops. Any line entering a closed surface must also exit it, resulting in a net flux of zero.
Consider a closed surface encompassing a bar magnet. According to Gauss's law for magnetism, the magnetic flux emerging from the North pole of the magnet will be:
Since the net magnetic flux through any closed surface is zero, any magnetic flux emerging from the North pole (considered positive) must be balanced by an equal amount of magnetic flux entering the South pole (considered negative) within the same closed surface. This is a direct consequence of magnetic field lines forming closed loops.
Which of the following is analogous to current in Ampere's circuital law, when comparing it to Gauss's law for electrostatics?
The NCERT text states: 'Ampere’s law is to Biot-Savart law, what Gauss’s law is to Coulomb’s law. Both, Ampere’s and Gauss’s law relate a physical quantity on the periphery or boundary (magnetic or electric field) to another physical quantity, namely, the source, in the interior (current or charge).' Here, current acts as the source for the magnetic field in Ampere's law, similar to how electric charge acts as the source for the electric field in Gauss's law for electrostatics.
A closed surface encloses an electric dipole. What is the net electric flux through the surface according to Gauss's law for electrostatics?
For an electric dipole, the total enclosed charge is $q + (-q) = 0$. According to Gauss's law for electrostatics ($\oint \vec{E} \cdot d\vec{A} = q_{\text{enclosed}}/\epsilon_0$), if the net enclosed charge is zero, the net electric flux through the closed surface is also zero.
Which of the following is true regarding magnetic field lines and electric field lines?
The text states: 'These lines called magnetic field lines form closed loops. This is unlike the electrostatic field lines which originate from positive charges and end at negative charges.' This is a key distinction between the two types of fields.
In the energy band diagram of an n-type silicon semiconductor at room temperature, where is the donor energy level ($E_D$) located relative to the conduction band ($E_C$)?
According to the NCERT text, 'In the energy band diagram of n-type Si semiconductor, the donor energy level $E_D$ is slightly below the bottom $E_C$ of the conduction band and electrons from this level move into the conduction band with very small supply of energy.' This placement facilitates electron donation to the conduction band.
For a p-type semiconductor, what is the location of the acceptor energy level ($E_A$) relative to the valence band ($E_V$)?
The NCERT states, 'for p-type semiconductor, the acceptor energy level $E_A$ is slightly above the top $E_V$ of the valence band... With very small supply of energy an electron from the valence band can jump to the level $E_A$ and ionise the acceptor negatively.' This creates holes in the valence band, which are majority carriers in a p-type semiconductor.
In n-type semiconductors at room temperature, what is the primary source of electrons in the conduction band?
The context mentions, 'at room temperature, most of the donor atoms get ionised but very few (~10^12) atoms of Si get ionised. So the conduction band will have most electrons coming from the donor impurities.' This highlights that doping is the dominant source of conduction electrons in n-type semiconductors.
At room temperature, in a p-type extrinsic semiconductor, what primarily determines the density of holes in the valence band?
The NCERT text explains, 'At room temperature, most of the acceptor atoms get ionised leaving holes in the valence band. Thus at room temperature the density of holes in the valence band is predominantly due to impurity in the extrinsic semiconductor.'
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