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Physics 4 Marks Questions

MODEL PAPER 7 · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 2 questions.

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Question 14 Marks
A stationary charge produces only an electrostatic field while a charge in uniform motion produces a magnetic field, that does not change with time. An oscillating charge is an example of accelerating charge. It produces an oscillating magnetic field, which in turn produces an oscillating electric fields and so on. The oscillating electric and magnetic fields regenerate each other as a wave which propagates through space.
Image
(i). Magnetic field in a plane electromagnetic wave is given by $\vec{B}= B _0 \sin ( kx +\omega t ) \hat{j} T$Expression for corresponding electric field will be (Where c is speed of light.)
(a) $\vec{E}= B _0 c \sin ( kx +\omega t ) \hat{k} V / m$
(b) $\vec{E}=- B _0 c \sin ( kx -\omega t ) \hat{k} V / m$
(c) $\vec{E}=- B _0 c \sin ( kx +\omega t ) \hat{k} V / m$
(d) $\vec{E}=\frac{B_0}{c} \sin ( kx +\omega t ) \hat{k} V / m$

(ii) The electric field component of a monochromatic radiation is given by $\vec{E}=2 E _0 \hat{i} \cos kz \cos \omega t$. Its magnetic field $\vec{B}$ is then given by
(a) $-\frac{2 E_0}{c} \hat{j} \sin kz \sin \omega t$
(b) $\frac{2 E_0}{c} \hat{j} \sin kz \sin \omega t$
(c) $\frac{2 E_0}{c} \hat{j} \sin kz \cos \omega t$
(d) $\frac{2 E_0}{c} \hat{j} \cos kz \cos \omega t$

(iii) A plane em wave of frequency 25 MHz travels in a free space along x -direction. At a particular point in space and time, $E =(6.3 \hat{j}) V / m$. What is magnetic field at that time?
(a) $0.089 \mu T$
(b) $0.124 \mu T$
(c) $0.021 \mu T$
(d) $0.095 \mu T$

OR

A plane electromagnetic wave travels in free space along $x$-axis. At a particular point in space, the electric field along $y$-axis is $9.3 V m ^{-1}$. The magnetic induction (B) along z -axis is
(a) $3.1 \times 10^{-8} T$
(b) $3 \times 10^{-5} T$
(c) $3 \times 10^{-6} T$
(d) $9.3 \times 10^{-6} T$

(iv) A plane electromagnetic wave travelling along the $x$-direction has a wavelength of 3 mm . The variation in the electric field occurs in the $y$-direction with an amptitude $66 V m ^{-1}$. The equations for the electric and magnetic fields as a function of $x$ and $t$ are respectively
$\begin{aligned} \text { a) } E_y & =11 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right), & \text { b) } E_y & =66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right), \\ B_y & =11 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) & B_z & =2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) \\ \text { c) } E_x & =33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right), & \text { d) } E_y & =33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right), \\ B_x & =11 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right) & B_z & =1.1 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\end{aligned}$
Answer
A stationary charge produces only an electrostatic field while a charge in uniform motion produces a magnetic field, that does not change with time. An oscillating charge is an example of accelerating charge. It produces an oscillating magnetic field, which in turn produces an oscillating electric fields and so on. The oscillating electric and magnetic fields regenerate each other as a wave which propagates through space. 
Image
(i)  (a) $\vec{E}= B _0 c \sin ( kx +\omega t ) \hat{k} V / m$
Explanation: Given : $\vec{B}= B _0 \sin ( kx +\omega t ) \hat{j} T$
The relation between electric and magnetic field is, $c =\frac{E}{B}$ or $E = cB$
The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along z-axis is obtained as $\vec{E}= cB _0 \sin ( kx +\omega t ) \hat{k} V / m$

(ii) (b) $\frac{2 E_0}{c} \hat{j} \sin kz \sin \omega t$
Explanation: $\frac{d E}{d z}=-\frac{d B}{d t}$
$\frac{d E}{d z}=-2 E _0 k \sin kz \cos \omega t =-\frac{d B}{d t}$
$dB =+2 E _0 k \sin kz \cos \omega tdt$
$B =+2 E _0 k \sin kz \int \cos \omega tdt =+2 E _0 \frac{k}{\omega} \sin kz \sin \omega t$ $\frac{E_0}{B_0}=\frac{\omega}{k}=c$
$B =\frac{2 E_0}{c} \sin kz \sin \omega t \therefore \vec{B}=\frac{2 E_0}{c} \sin kz \sin \omega t \hat{j}$
$E$ is along $y$-direction and the wave propagates along $x$-axis.
$\therefore B$ should be in a direction perpendicular to both $x$-and $y$-axis.

(iii) (c) $0.021 \mu T$
Explanation: Here, $E =6.3 \hat{j} ; c =3 \times 10^8 m / s$
The magnitude of $B$ is
$B_{Z}=\frac{E}{c}=\frac{6.3}{3 \times 10^8}=2.1 \times 10^{-8} T=0.021 \mu T$

OR

(a) $3.1 \times 10^{-8} T$
Explanation: At a particular point, $E =9.3 V m ^{-1}$
$\therefore$ Magnetic field at the same point $=\frac{9.3}{3 \times 10^8}$
$=3.1 \times 10^{-8} T$

(iv) (b) $E_y=66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right), B_z=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$
Explanation: Here : $E _0=66 Vm ^{-1}, E _y=66 \cos \omega\left(t-\frac{x}{c}\right)$,
$\begin{array}{l}\lambda=3 mm=3 \times 10^{-3} m, k=\frac{2 \pi}{\lambda} \\\frac{\omega}{k}=c \Rightarrow \omega=ck=3 \times 10^8 \times \frac{2 \pi}{3 \times 10^{-3}} \\\text { or } \omega=2 \pi \times 10^{11}\end{array}$
$\begin{array}{l}\therefore E _{ y }=66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) \\ 
B _{ z }=\frac{E_y}{c}=\left(\frac{66}{3 \times 10^8}\right) \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) \\ 
=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\end{array}$

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Question 24 Marks
Net electric flux through a cube is the sum of fluxes through its six faces. Consider a cube as shown in figure, having sides of length $L =10.0 cm$. The electric field is uniform, has a magnitude $E =4.00 \times 10^3 NC ^{-1}$ and is parallel to the xy plane at an angle of $37^{\circ}$ measured from the + x -axis towards the + y -axis.
Image
(i) Electric flux passing through surface $S_6$ is
(a) $-24 Nm ^2 C ^{-1}$
(b) $32 Nm ^2 C ^{-1}$
(c) $-32 Nm ^2 C ^{-1}$
(d) $24 Nm ^2 C ^{-1}$

(ii) Electric flux passing through surface $S_1$ is
(a) $-32 Nm ^2 C ^{-1}$
(b) $-24 Nm ^2 C ^{-1}$
(c) $32 Nm ^2 C ^{-1}$
(d) $24 Nm ^2 C ^{-1}$

(iii) The surfaces that have zero flux are
 (a) $S _2$ and $S _4$ (b) $S_3$ and $S_6$ (c) $S _1$ and $S _2$ (d) $S _1$ and $S _3$

(iv) he total net electric flux through all faces of the cube is a) 24 Nm2 C-1 c)-8 Nm2 C-1 b
(a) $24 Nm ^2 C ^{-1}$ (b) $8 Nm ^2 C ^{-1}$ (c) $-8 Nm ^2 C ^{-1}$ (d) zero

OR

The dimensional formula of surface integral $\oint \vec{E} \cdot d \vec{S}$ of an electric field is
(a) $\left[ M ^{-1} L^3 T^{-3} A\right]$
(b) $\left[ M L ^2 T^{-2} A^{-1}\right]$
(c) $\left[ ML ^3 T^{-3} A^{-1}\right]$
(d) $\left[ M L ^{-3} T^{-3} A^{-1}\right]$
Answer
Net electric flux through a cube is the sum of fluxes through its six faces. Consider a cube as shown in figure, having sides of length $L =10.0 cm$. The electric field is uniform, has a magnitude $E =4.00 \times 10^3 NC ^{-1}$ and is parallel to the xy plane at an angle of $37^{\circ}$ measured from the $+x$-axis towards the $+y$-axis.
Image
(i) (c) $-32 Nm ^2 C ^{-1}$
Explanation: Electric flux, $\phi=\vec{E} \cdot \vec{A}=E A \cos \theta$ 
where $\vec{A}=A \hat{n}$
For electric flux passing through $S _6, \hat{n}_{S_6}=-\hat{i}$ (Back)
$\begin{array}{l}\therefore \phi_{S_6}=-\left(4 \times 10^3 NC^{-1}\right)(0.1 m)^2 \cos 37^0 \\=-32 Nm^2 C^{-1}\end{array}$

(ii) (b) $-24 Nm ^2 C ^{-1}$
Explanation: For electric flux passing through $S _1, \hat{n}_{S_1}=-\hat{j}$ (Left)
$\begin{array}{l}\therefore \phi_{S_1}=-\left(4 \times 10^3 NC^{-1}\right)(0.1 m)^2 \cos 90^{\circ}=0 \\=-24 Nm^2 C^{-1}\end{array}$

(iii) (a) $S _2$ and $S _4$
Explanation: Here, $\hat{n}_{S_2}=+\hat{k}$ (Top)
$\begin{array}{l}\therefore \phi_{S_2}=-\left(4 \times 10^3 NC ^{-1}\right)(0.1 m)^2 \cos 90^{\circ}=0 \\ \hat{n}_{S_3}=+\hat{j} \text { (Right) } \\ \hat{n}_{S_4}=-\hat{k} \text { (Bottom) }\end{array}$
$\begin{array}{l}\therefore \phi_{S_4}=-\left(4 \times 10^3 NC ^{-1}\right)(0.1 m)^2 \cos 90^{\circ}=0 \\ \text { And, } \hat{n}_{S_5}=+\hat{i} \text { (Front) }\end{array}$
$ \begin{array}{l} \therefore \phi_{S_5}=+\left(4 \times 10^3 NC^{-1}\right)(0.1 m)^2 \cos 37^0 \\ =32 Nm^2 C^{-1} \end{array}$
$S_2$ and $S_4$ surface have zero flux.

(iv) (d) zero
Explanation: As the field is uniform, the total flux through the cube must be zero, i.e., any flux entering the cube must leave it.

OR

(c) $\left[ M L ^3 T^{-3} A^{-1}\right]$
Explanation: Surface integral $\oint \vec{E} \cdot d \vec{S}$ is the net electric flux over a closed surface S.
$\therefore\left[\phi_E\right]=\left[ M L ^3 T^{-3} A^{-1}\right]$

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