Q: The dimension of \$1/2\varepsilon_oE^2\$(\$\varepsilon_o\$ permittivity of free space; E: electric field) is?

Dimension of \$E_0=M^{-1}L^{-3}T^4A^2\$Dimension of E(Electric field)=\$MLT^{-2}/AT\\=MLT^{-1}A^{-1}\$Dimension of \$\cfrac{1}{2\varepsilon _0 E^2}=\cfrac{1}{(M^{-1}L^{-3}T^4A^2)(MLT^{-1}A^{-1})^2}\\=\cfrac{1}{ML^{-1}T^2}=ML^{-1}T^{-2}\$

Q: \$F_g\$ and \$F_e\$ represent gravitational and electrostatic force respectively between electrons situated at a distance \$0.1\,m. F_g/F_e\$ is of the order

\${ F }_{ y }\$ and \${ F }_{ e }\$ represent gravitational and electrostatic force respectively between electrons situated at a distance \$=0.1m\$let, us take the ratio\$\dfrac { { F }_{ e } }{ { F }_{ g } } =\dfrac { \dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }^{ 2 } }{ { d }^{ 2 } } }{ G\dfrac { { m }^{ 2 } }{ { d }^{ 2 } } } \$\$\dfrac { { F }_{ e } }{ { F }_{ g } } =\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \times \dfrac { 1 }{ G } \times \dfrac { { q }^{ 2 } }{ { m }^{ 2 } } \$\$\dfrac { { F }_{ e } }{ { F }_{ g } } =9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 6.67\times { 10 }^{ -11 } } \times \dfrac { { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ { \left( 9.1\times { 10 }^{ -31 } \right) }^{ 2 } } \$\$\Rightarrow \dfrac { { F }_{ e } }{ { F }_{ g } } =4.17\times { 10 }^{ 42 }\$or, \$\dfrac { { F }_{ e } }{ { F }_{ g } } \$ is of the order of \${ 10 }^{ -42 }\$.

Q: The ratio of magnitude of electrostatic force and gravitational force for an electron and proton is

Here, for an electron and a protonElectrostatic force, \$\left| { F }_{ e } \right| =\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { e }^{ 2 } }{ { r }^{ 2 } } \$Gravitational force, \$\left| { F }_{ g } \right| =\cfrac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } \$\$\therefore \cfrac { \left| { F }_{ e } \right| }{ \left| { F }_{ g } \right| } =\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { e }^{ 2 } }{ G{ m }_{ e }{ m } } =\cfrac { 9\times { 10 }^{ 9 }\times { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ 6.67\times { 10 }^{ -11 }\times 9\times { 10 }^{ -31 }\times 1.66\times { 10 }^{ -27 } } =2.4\times { 10 }^{ 39 }\$

Q: In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is \$0.26\ G\$ and the angle of dip is \$60^o\$. The magnetic field of the earth at this location is

The earth's magnetic field is \$B_e\$ and its horizontal and vertical components are \$H_e\$ and \$H_v\$\$\cos \theta =\dfrac{H_e}{B_e}\$\$ \therefore \cos 60^{\circ} = \dfrac{0.26 \times 10^{-4}}{B_e}\:T\$\$ \Rightarrow B_e = \dfrac{ 0.26 \times 10^{-4}}{ \dfrac{1}{2}}=0.52 \times 10^{-4}\:T=0.52\:G\$

Question 1

Why do magnetic field lines not intersect each other?

Accepted Answer

If magnetic field lines intersect each other, then at the intersection point there will be two directions of the same field which is not possible. Hence the field lines do not cross or intersect each other.

Question 2

The magnetic field in a given region is uniform. Draw a diagram to represent it.

Accepted Answer

Uniform magnetic field is  represented by  parallel magnetic field lines  in same direction.

Question 3

The dimension of $1/2\varepsilon_oE^2$($\varepsilon_o$ permittivity of free space; E: electric field) is?

Accepted Answer

Dimension of $E_0=M^{-1}L^{-3}T^4A^2$Dimension of E(Electric field)=$MLT^{-2}/AT\=MLT^{-1}A^{-1}$Dimension of $\cfrac{1}{2\varepsilon _0 E^2}=\cfrac{1}{(M^{-1}L^{-3}T^4A^2)(MLT^{-1}A^{-1})^2}\=\cfrac{1}{ML^{-1}T^2}=ML^{-1}T^{-2}$

Question 4

$F_g$ and $F_e$ represent gravitational and electrostatic force respectively between electrons situated at a distance $0.1\,m. F_g/F_e$ is of the order

Accepted Answer

${ F }_{ y }$ and ${ F }_{ e }$ represent gravitational and electrostatic force respectively between electrons situated at a distance $=0.1m$let, us take the ratio$\dfrac { { F }_{ e } }{ { F }_{ g } } =\dfrac { \dfrac { 1 }{ 4\pi { \epsilon  }_{ 0 } } \dfrac { { q }^{ 2 } }{ { d }^{ 2 } }  }{ G\dfrac { { m }^{ 2 } }{ { d }^{ 2 } }  } $$\dfrac { { F }_{ e } }{ { F }_{ g } } =\dfrac { 1 }{ 4\pi { \epsilon  }_{ 0 } } \times \dfrac { 1 }{ G } \times \dfrac { { q }^{ 2 } }{ { m }^{ 2 } } $$\dfrac { { F }_{ e } }{ { F }_{ g } } =9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 6.67\times { 10 }^{ -11 } } \times \dfrac { { \left( 1.6\times { 10 }^{ -19 } \right)  }^{ 2 } }{ { \left( 9.1\times { 10 }^{ -31 } \right)  }^{ 2 } } $$\Rightarrow \dfrac { { F }_{ e } }{ { F }_{ g } } =4.17\times { 10 }^{ 42 }$or,  $\dfrac { { F }_{ e } }{ { F }_{ g } } $ is of the order of ${ 10 }^{ -42 }$.

Question 5

The ratio of magnitude of electrostatic force and gravitational force for an electron and proton is

Accepted Answer

Here, for an electron and a protonElectrostatic force, $\left| { F }_{ e } \right| =\cfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \cfrac { { e }^{ 2 } }{ { r }^{ 2 } } $Gravitational force, $\left| { F }_{ g } \right| =\cfrac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } $$\therefore \cfrac { \left| { F }_{ e } \right|  }{ \left| { F }_{ g } \right|  } =\cfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \cfrac { { e }^{ 2 } }{ G{ m }_{ e }{ m } } =\cfrac { 9\times { 10 }^{ 9 }\times { \left( 1.6\times { 10 }^{ -19 } \right)  }^{ 2 } }{ 6.67\times { 10 }^{ -11 }\times 9\times { 10 }^{ -31 }\times 1.66\times { 10 }^{ -27 } } =2.4\times { 10 }^{ 39 }$

Question 6

Identify the poles of the magnet in the figure (a) and (b) shown below.

Accepted Answer

The magnetic field lines emerge from north pole and merge at the south pole

Question 7

In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $0.26\  G$ and the angle of dip is $60^o$. The magnetic field of the earth at this location is

Accepted Answer

The earth's magnetic field is $B_e$ and its horizontal and vertical components are $H_e$ and $H_v$$\cos \theta =\dfrac{H_e}{B_e}$$ \therefore \cos 60^{\circ} = \dfrac{0.26 \times 10^{-4}}{B_e}\:T$$ \Rightarrow B_e = \dfrac{ 0.26 \times 10^{-4}}{ \dfrac{1}{2}}=0.52 \times 10^{-4}\:T=0.52\:G$

Question 8

Explain the magnetic elements of the earth's magnetic field.

Accepted Answer

The elements are:Magnetic declination (q)Magnetic inclination (d)Horizontal component (BH)&#8203;Magnetic declination:The small angle between magnetic axis and geographic axis at a place is defined as magnetic declination at that place. It is represented by &#952;.Magnetic Inclination:Magnetic dip, dip angle, or magnetic inclination is the angle made with the horizontal by the compass needle of a vertically held compass. This angle varies at different points on the Earth's surface. Positive values of inclination indicate that the magnetic field of the Earth is pointing downward, into the Earth, at the point of measurement. The value can be measured with an instrument typically known as a dip circle.(1)At equator the magnetic field is parallel to the horizon so there will be zero angle of dip.(2) At poles the magnetic field is almost vertical, here the dip will me maximum.Horizontal component :It is the component of earth&#8217;s magnetic field along the horizontal direction.So the angle of dip is the angle between earth&#8217;s magnetic field i.e. B and the horizontal component i.e. BH.

Question 9

Choose the incorrect statement from the following:

Accepted Answer

Magnetic field lines are the curved lines emanating from north pole to south pole of a magnet tangent at every point of the field lines represents the direction magnetic field at that point. So, they can never intersect otherwise there will be more than one directions of magnetic field represented by the tangents at the point of intersection.Option C is correct.

Question 10

The substances which are attracted towards a magnet are called _______.

Accepted Answer

The substances which are attracted towards a magnet are called ferro-magnetic materials Materials that can be magnetized, which are also the ones that are strongly attracted to a magnet, are called ferromagnetic (or ferrimagnetic). These include iron, nickel, cobalt, some alloys of rare-earth metals, and some naturally occurring minerals such as lodestone.