Class 12th

1.31 A proton has 3 quarks. Let there be n 'up quarks', then number of 'down quarks' = 3-n

Charge due to n 'up quarks' = $(\frac{2}{3}$ e) n

Charge due to (3-n) 'down quarks' =$-$$\frac{1}{3}e\left)\right(3-n)$ 

Total charge on a proton = +e = $(\frac{2}{3}$ e) n +  ( -$\frac{1}{3}e\left)\right(3-n)$ 

e =  $\frac{2ne}{3}$ +$\frac{ne}{3}$-e

2e = $\frac{3ne}{3}$

n = 2

Number of 'up quark' = 2 and number of 'down quark' = 1. Therefore a proton can be represented as 'uud'.

A neutron has 3 quarks. Let there be n 'up quark' in a neutron and (3-n) 'down quark'

Charge due to n 'up quark' = +$(\frac{2}{3}$ e)n

Charge due to (3-n) 'down quark' = 
- (

Since total charge of a neutron is zero, we get

+ $(\frac{2}{3}$e)n -($\frac{1}{3}e\left)\right(3-n)$ = 0

$(\frac{2}{3}$ e)n = ( $\frac{1}{3}e\left)\right(3-n)$

en = e or n = 1

Hence number of 'up quark' in neutron = 1

Number of 'down quark' in neutron = 3-1 = 2

Therefore a neutron can be represented as 'udd'.

1.30 

Take a long thin wire XY of uniform linear charge density $\lambda$. Consider a point A at a perpendicular distance l from the midpoint O of the wire. Let E be the electric field at point A due to the wire XY. Consider a small length element dx on the wire section with OZ = x. Let q be the charge on this piece.

Q = $\lambda dx$

Electric field due to the piece,

dE =  = $\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda dx}{{\left(AZ\right)}^2}$  = $\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda dx}{(l^2+x^2)}$since AZ =$\text{exception 25:}$

The electric field is resolved into two rectangular components. dE $\cos ?\theta$ is the perpendicular component and dE$\sin ?\theta$ $$ When the whole wire is considered, the component dE $\sin ?\theta \mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{e}\mathrm{l}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}.\mathrm{}$ is cancelled. Only the perpendicular component dE $\cos ?\theta$ affects point A.

Hence, effective electric field at point A due to element dx is$d{E}_1$ 

 ………………….(1)

In  , $?AZO$ ,  $\tan ?\theta =\frac{x}{l}$ ,x =$l\tan ?\theta$……….(2)

On differentiating equation (2), we get

$\frac{dx}{d\theta }=l{sec}^2\theta$ , dx = …………….(3)

From equation (2), we have

 $x^2$ +$l^2$ = $l^2{tan}^2\theta +l^2$ =$l^2{sec}^2\theta$…….(4)

Putting equations (3) and (4) in equation (1), we get

 $d{E}_1=\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda l{sec}^2\theta d\theta \cos ?\theta }{l^2{sec}^2\theta }$ =$\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda \cos ?\theta d\theta }{l}$…….(5)

The wire is so long that $\theta$tends from  $-\frac{\pi }{2}$to$\frac{\pi }{2}$

By integrating equation (5), we get

 ${\int }_{-\frac{\pi }{2}}^{\mathrm{}\frac{\pi }{2}}d{E}_{\mathrm{1=}}$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda \cos ?\theta d\theta }{l})$

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{l}{\left[\sin ?\theta \right]}_{\mathrm{}-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{l}*2=\frac{\lambda }{2\pi {\epsilon }_{0}l}$

Therefore the electric field due to long wire is $\frac{\lambda }{2\pi {\epsilon }_{0}l}$

8.8 Electric field amplitude, $E_0$ = 120 N/C

Frequency of source, $\nu$ = 50 MHz = 50 $*{10}^6$ Hz

Speed of light, c = 3 $*{10}^8$ m/s

Magnitude of magnetic field strength is given as

$B_0=$ $\frac{E_0}{c}$ = $\frac{120}{3\mathrm{}*{10}^8}$ = 4 $*{10}^{-7}$ T = 400 $*{10}^{-9}$ T = 400 nT

$\mathrm{A}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{a}\mathrm{s}\mathrm{}$ $\omega$ = 2 $\pi \nu$ = 2 $*\pi *$ 50 $*{10}^6$

= 3.14 $*{10}^8$ rad/s

Propagation constant is given as

k = $\frac{\omega }{c}$ = $\frac{3.14\mathrm{}*{10}^8}{3\mathrm{}*{10}^8}$ = 1.05 rad/m

Wavelength of wave is given as

$\lambda =\frac{c}{\nu }$ = $\frac{3\mathrm{}*{10}^8}{50\mathrm{}*{10}^6}$ = 6.0 m

Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

$\stackrel{?}{E}$ = $E_0\sin ?(kx-\omega t)\stackrel{?}{j}$

$=$ 120 $*\mathrm{s}\mathrm{i}\mathrm{n}?(1.05*x-$ 3.14 $*{10}^{8}t)\stackrel{?}{j}$

Equation of magnetic field vector is given as:

$\stackrel{?}{B}$ = $B_0\sin ?(kx-\omega t)\stackrel{?}{k}$

= (400 $*{10}^{-9})\sin ?(1.05*x-$ 3.14 $*{10}^{8}t$ ) $\stackrel{?}{k}$

1.29 Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E be the electric field outside the conductor, q is the electric charge, $\sigma$is the charge density and ${\epsilon }_0$ is the permittivity of free space.

Charge q =$\sigma$ $*ds$ 

According to Gauss's law, flux$\phi$ = E.ds = $\frac{q}{{?}_0}$=$\frac{\sigma \mathrm{}*ds}{{?}_0}$

Hence, E = $\frac{\sigma }{2{\epsilon }_0}\stackrel{}{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{{2\epsilon }_0}\stackrel{}{n}$ . This field is a superposition of field due to the cavity E' and the field due to the rest of the charged conductor E'. These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor.

So E' + E' = E

E'  = $\frac{E}{2}$=$\frac{\sigma }{2{\epsilon }_0}\stackrel{}{n}$

Hence, the field due to the rest of the conductor is $\frac{\sigma }{2{\epsilon }_0}\stackrel{}{n}$
$\frac{}{}\stackrel{}{?}$