Physics Ncert Solutions Class 12th

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 $\frac{\sigma }{2{\epsilon }_0}\stackrel{}{n}$
$\frac{}{}\stackrel{}{?}$

1.28 (a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.

Let q be the charge inside the conductor and ${?}_0$ the permittivity of free space.

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

Here, E = 0, hence $\frac{q}{{?}_0}$ = 0, so q = 0 (as ${?}_0\ne 0)$

Therefore, the charge inside the conductor is zero. The entire charge Q appears on the outer surface of the conductor.

(b) The outer surface of the conductor A has a charge amount Q. Another conductor B, having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount –q will be induced in the inner surface of the conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of the conductor A is Q+q.

(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.

8.2 Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of parallel capacitor, C = 100 pF = 100 $*{10}^{-12}$ F

Supply voltage, V = 230 V

Angular frequency, $\omega$ = 300 rad/s

rms value of the conduction current, I = $\frac{V}{X_c}$ , where $X_c=\mathrm{c}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}=\mathrm{}\frac{1}{\mathrm{\omega }\mathrm{C}}$

Hence, I = V $*\omega *C$ = 230 $*300*$ 100 $*{10}^{-12}$ = 6.9 $*{10}^{-6}$ A = 6.9 $\mu A$

Yes, the conduction current is equal to the displacement current.

Magnetic field is given as, B = $\frac{{\mu }_{0}r}{2\pi R^2}{I}_0$ , where

${\mu }_0$ = Free space permeability = 4 $\pi *{10}^{-7}$ N $A^{-2}$

$I_0$ = Maximum value of current = $\surd 2I$

r = distance between two plates on the axis = 3.0 cm = 0.03 m

then, B = $\frac{4\pi *{10}^{-7}*0.03}{2\pi *{0.06}^2}$ $*\surd 2*$ 6.9 $*{10}^{-6}$ = 1.626 $*{10}^{-11}$ T