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: \$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: 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)

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.

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

$F_g$ 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

The given magnet is divided into three parts A, B and C.

A B C
Name the parts where the strength of the magnetic field is:
(i) maximum (ii) minimum
How will density of magnetic field lines differ at these parts.

Accepted Answer

(i) Maximum magnetic field is in the region A and C. (ii) Minimum magnetic field is in the region B. Reason: A and C are magnetic poles and have maximum number of magnetic field lines which determine the intensity of magnetic field while B is centre of the magnet has no magnetic field lines. So intensity of magnetic field near B is almost zero.

Question 8

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 9

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 10

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.

Electrostatic, magnetic and gravitational force

Language

Rate

REVISE WITH CONCEPTS

Contact and Non-Contact Forces

Non-Contact Forces

Compare Electric, Magnetic Fields With Gravitational Fields

QUICK SUMMARY WITH STORIES