Q: Derive an expression for magnetic field at a point on the axial line of a bar magnet.

Let \$NS\$ be a bar magnet of magnetic length \$2l\$ and having each pole of magnetic strength \$m\$ . O is the center of magnet and \$P\$ is a point on axial line at a distance \$r\$ from the center \$O\$ of magnet , at which magnetic field has to be measured .The magnetic field \$B_{1}\$ at \$P\$ due to \$N\$ pole of magnet , \$B_{1}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{NP^{2}}\$or \$B_{1}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r-l)^{2}}\$ (along PX) .......................eq1And , the magnetic field \$B_{2}\$ at \$P\$ due to \$S\$ pole of magnet , \$B_{2}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{SP^{2}}\$or \$B_{2}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r+l)^{2}}\$ (along PS) ......................eq2Therefore , resultant magnetic field at point \$P\$ , \$B=B_{1}-B_{2}\$ (-ive sign is due to opposite directionS of \$B_{1}\$ and \$B_{2}\$)It is clear from eq1 and eq2 that \$B_{1}>B_{2}\$ ,therefore the direction of \$B\$ will be along \$PX\$ .or \$B=\$\$\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r-l)^{2}}\$\$-\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r+l)^{2}}\$ (alongPX)or \$B=\dfrac{\mu_{0}}{4\pi}.\dfrac{m(4rl)}{(r^{2}-l^{2})^{2}}\$ (alongPX)Now , \$m(2l)=M\$ (magnetic dipole moment of magnet)Hence , \$B=\dfrac{\mu_{0}}{4\pi}.\dfrac{2Mr}{(r^{2}-l^{2})^{2}}\$ (alongPX)

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: \$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: Check that the ratio \$ke^2\$ /G \$m_e\,m_p\$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

The given ratio is \$\displaystyle \frac{ke^2}{Gm_cm_p}\$Where,\$G =\$ Gravitational constantIts unit is \$N\,m^2kg^{-2}\$.\$m_e\$ and \$m_p=\$ Masses of electron and proton.Their unit is kg.\$e =\$ Electric charge.Its unit is C.\$k = A\$ constant \$\displaystyle =\frac{1}{4\pi\in_0}\$\$\in_0=\$ Permittivity of free spaceIts unit is \$N\,m^2C^{-2}\$. Unit of the given ratio \$\displaystyle \frac{ke^2}{Gm_cm_p}=\frac{[Nm^2C^2][C^{-2}]}{[Nm^2\,kg^{-2}][kg][kg]}\$ \$=M^0L^0T^0\$Hence, the given ratio is dimensionless.\$e=1.6\times 10^{-19}C\$\$G=6.67\times 10^{-11}N\,m^2kg^{-2}\$\$m_e=9.1\times 106{-31}kg\$\$m_p=1.66\times 10^{-27}kg\$\$\displaystyle \frac{ke^2}{Gm_cm_p} = \frac{9\times 10^9\times(1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-3}\times 1.67\times 10^{-22}}\approx 2.3\times 10^{39}\$This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Q: The ratio of electric force to the gravitational force between two protons separated by a finite distance is of the order of:

\$Gravitational\ force\ = \dfrac{G(m_p)^2}{r^2}\$\$Electric\ force\ = \dfrac{K(e)(e)}{r^2}\$\$Ratio\ = \dfrac{K(1.6\times 10^{-19})^2}{r^2}\times \dfrac{r^2}{G\times (1.6\ 7\times 10^{-27})^2}\$\$= \dfrac{(9\times 10^9)\times (1.6\times 10^{-19})^2}{(6.6\ 7\times 10^{-11})(1.6\ 7\times 10^{-27})^2}\$\$=1.23\times 10^{20}\times 10^{16}\$\$= 10^{36}\$(order of magnitude)

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

Derive an expression for magnetic field at a point on the axial line of a bar magnet.

Accepted Answer

Let $NS$ be a bar magnet of magnetic length $2l$ and having each pole of magnetic strength $m$ . O is the center of magnet and $P$ is a point on axial line at a distance $r$ from the center $O$ of magnet , at which magnetic field has to be measured .The magnetic field $B_{1}$ at $P$ due to $N$ pole of magnet ,      $B_{1}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{NP^{2}}$or   $B_{1}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r-l)^{2}}$   (along PX) .......................eq1And , the magnetic field $B_{2}$ at $P$ due to $S$ pole of magnet ,      $B_{2}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{SP^{2}}$or   $B_{2}=\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r+l)^{2}}$   (along PS) ......................eq2Therefore , resultant magnetic field at point $P$ ,       $B=B_{1}-B_{2}$  (-ive sign is due to opposite directionS of $B_{1}$ and $B_{2}$)It is clear from eq1 and eq2 that $B_{1}>B_{2}$ ,therefore the direction of  $B$ will be along $PX$ .or    $B=$$\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r-l)^{2}}$$-\dfrac{\mu_{0}}{4\pi}.\dfrac{m}{(r+l)^{2}}$     (alongPX)or    $B=\dfrac{\mu_{0}}{4\pi}.\dfrac{m(4rl)}{(r^{2}-l^{2})^{2}}$     (alongPX)Now , $m(2l)=M$ (magnetic dipole moment of magnet)Hence , $B=\dfrac{\mu_{0}}{4\pi}.\dfrac{2Mr}{(r^{2}-l^{2})^{2}}$     (alongPX)

Question 4

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 5

$F_g$ 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 6

Check that the ratio $ke^2$ /G $m_e\,m_p$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Accepted Answer

The given ratio is $\displaystyle \frac{ke^2}{Gm_cm_p}$Where,$G =$ Gravitational constantIts unit is $N\,m^2kg^{-2}$.$m_e$ and $m_p=$ Masses of electron and proton.Their unit is kg.$e =$ Electric charge.Its unit is C.$k = A$ constant   $\displaystyle =\frac{1}{4\pi\in_0}$$\in_0=$ Permittivity of free spaceIts unit is $N\,m^2C^{-2}$. Unit of the given ratio $\displaystyle \frac{ke^2}{Gm_cm_p}=\frac{[Nm^2C^2][C^{-2}]}{[Nm^2\,kg^{-2}][kg][kg]}$                                                                        $=M^0L^0T^0$Hence, the given ratio is dimensionless.$e=1.6\times 10^{-19}C$$G=6.67\times 10^{-11}N\,m^2kg^{-2}$$m_e=9.1\times 106{-31}kg$$m_p=1.66\times 10^{-27}kg$$\displaystyle \frac{ke^2}{Gm_cm_p} = \frac{9\times 10^9\times(1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-3}\times 1.67\times 10^{-22}}\approx 2.3\times 10^{39}$This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Question 7

The ratio of electric force to the gravitational force between two protons separated by a finite distance is of the order of:

Accepted Answer

$Gravitational\ force\ = \dfrac{G(m_p)^2}{r^2}$$Electric\ force\ = \dfrac{K(e)(e)}{r^2}$$Ratio\ = \dfrac{K(1.6\times 10^{-19})^2}{r^2}\times \dfrac{r^2}{G\times (1.6\ 7\times 10^{-27})^2}$$= \dfrac{(9\times 10^9)\times (1.6\times 10^{-19})^2}{(6.6\ 7\times 10^{-11})(1.6\ 7\times 10^{-27})^2}$$=1.23\times 10^{20}\times 10^{16}$$= 10^{36}$(order of magnitude)

Question 8

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 9

What are magnetic field lines? How is the direction of magnetic field at a point in a magnetic field determined using field lines?

Accepted Answer

A magnetic field line is the path along which a free north-pole tends to move. The direction of a magnetic field at a point is determined by placing a small compass needle. The N - pole of compass indicates the direction of magnetic field at that point.

Question 10

Draw a diagram to show the magnetic fields lines around a bar magnet.

Accepted Answer

Magnetic fields lines around a bar magnet.

Introduction to Magnetic Field and Magnetic Field Lines

Rate

REVISE WITH CONCEPTS

Recall Basics of Magnets

Nature of Magnetic Field and Field Lines

Properties of Magnetic Field Lines Around a Bar Magnet

Tracing Magnetic Field

QUICK SUMMARY WITH STORIES

Magnetic Field

Nature of Magnetic Field and Field Lines

Properties of Magnetic Field Lines Around a Bar Magnet

Introduction to Magnetic Field and Magnetic lines

Why do magnetic field lines not intersect each other?

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

Derive an expression for magnetic field at a point on the axial line of a bar magnet.

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

$F_{g}$ 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

Check that the ratio $k e^{2}$ /G $m_{e} m_{p}$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

The ratio of electric force to the gravitational force between two protons separated by a finite distance is of the order of:

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

What are magnetic field lines? How is the direction of magnetic field at a point in a magnetic field determined using field lines?

Draw a diagram to show the magnetic fields lines around a bar magnet.

Introduction to Magnetic Field and Magnetic Field Lines

Rate

REVISE WITH CONCEPTS

Recall Basics of Magnets

Nature of Magnetic Field and Field Lines

Properties of Magnetic Field Lines Around a Bar Magnet

Tracing Magnetic Field

QUICK SUMMARY WITH STORIES

Magnetic Field

Nature of Magnetic Field and Field Lines

Properties of Magnetic Field Lines Around a Bar Magnet

Introduction to Magnetic Field and Magnetic lines

Why do magnetic field lines not intersect each other?

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

Derive an expression for magnetic field at a point on the axial line of a bar magnet.

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

Fg​ and Fe​ represent gravitational and electrostatic force respectively between electrons situated at a distance 0.1m.Fg​/Fe​ is of the order

Check that the ratio ke2 /G me​mp​ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

The ratio of electric force to the gravitational force between two protons separated by a finite distance is of the order of:

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

What are magnetic field lines? How is the direction of magnetic field at a point in a magnetic field determined using field lines?

Draw a diagram to show the magnetic fields lines around a bar magnet.

$F_{g}$ 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

Check that the ratio $k e^{2}$ /G $m_{e} m_{p}$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?