For coil x
Radius of coil, r_x = 16 cm = 0.16 m
Number of turns, n_x = 20
Current in the coil, l_x = 16 A (anti=clockwise)
                        

For coil y
Radius of coil, r_y = 10 cm = 0.1 m
Number of turns, n_y = 25
Current in the coil, I_y = 18 A (clockwise)
The magnitude of the magnetic field at the centre of coil x,
B_x = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2{ I }_{ x }\pi { n }_{ x } }{ { r }_{ x } } =\frac { { 10 }^{ -7 }\times 2\times 16\times \pi \times 20 }{ 0.16 } \)
= \(4\pi \times { 10 }^{ -4 }T\) 
The direction of magnetic field due to the coil x at centre O is towards right, i.e. East, according to right hand thumb rule. The magnitude of the magnetic field at the centre of coil y,
B_y = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2\pi { I }_{ y }{ n }_{ y } }{ { r }_{ y } } =\frac { { 10 }^{ -7 }\times 2\times \pi \times 18\times 25 }{ 0.1 } \)
= \(9\pi \times { 10 }^{ -4 }T\)
The direction of magnetic field due to coil y at centre O is towards left, i.e. West, according to right hand thumb rule. Here, the magnitude of B_y is greater than B_x, so the resultant magnetic field will be in the direction of B_y , i.e. left (West).  
Net magnetic field at the centre,
B = B_y - B_x = (9\(\pi\) - 4\(\pi\)) 10^-4 T = 5\(\pi\) \(\times\) 10^-4 T
(\(\because\) B_y and B_x are opposite to each other)
= 1.6 \(\times\) 10^-3 T (towards West)

(ii) As, the number of point sources increases, their contribution towards intensity also increases. Intensity varies as square of the slit width. Thus, when the width of the slit is made double the original width, intensity will get four times of its original value.
Width of central maximum is given by \(\beta =\frac { 2D\lambda }{ b } \)
Where, D = distance between screen and slit,
 \(\lambda \) = wavelength of the light,b = size of slit.

So, with the increase in size of slit, the width of central maxima decreases. Hence, double the size of the slit would result as half the width of the central maxima.

A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.

Hence, the output can be written as:
\(Y=\overline{A . A}=\bar{A}+\bar{A}=\bar{A}\)
The truth table for equation (i) can be drawn as:

 This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:

(i) The device that is used to bring high voltage down to low voltage of an AC current is a transformer. It works on the principle of mutual induction of two windings or circuits. When current in one circuit changes, emf is induced in the neighbouring circuit.
(ii) The transformer cannot convert DC voltages because it works on the principle of mutual induction. When the current linked with the primary coil changes, the magnetic flux linked with secondary coil also changes. This change in flux induces emf in the secondary coil. If we apply a direct current to the primary coil, then current will remain constant. Thus there is no mutual induction, and hence, no emf is induced.
(iii) The value of gaining knowledge and curiosity about learning new things is being displayed by the students. The value of providing good education and undertaking the doubts of students has been displayed by teacher.

i) The values developed by Mndeep were sympathy, responsibility, helping nature and self-reliance.
ii) The vlue imparted by mother are appreciation and thankfulness.
iii) It is not necessary that every magnetic configuration has a North pole and a South pole.It is true only if the source of the field has a net non-zero magnetic moment.This is not possible for a toroid or even for a straight infinite conductor.

Here R = 2m,
\(f=\frac { R }{ 2 } =\frac { 2 }{ 2 } =1m\)
Using morror formula, we have
\(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } \Rightarrow \frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u }\)
\( \Rightarrow \frac { 1 }{ v } =\frac { u-f }{ fu } \Rightarrow v=\frac { fu }{ u-f } \)
When Jogger is 39 m away, then u = -39m
\(\Rightarrow \) we get
\(v=\frac { fu }{ u-f } =\frac { 1\left( -39 \right) }{ -39-1 } or \ v=\frac { 39 }{ 40 } m\)
As, the Jogger is running at a constant speed of 5 m/s, after 1 s, the positionof the image(v) for
u = -39 + 5
u = -34m
Again using the Equation we get
\(\Rightarrow v=\frac { fu }{ u-f } =\frac { 1\left( -34 \right) }{ -34-1 } or\quad v=\frac { 34 }{ 35 } m\)
Difference in apparent position of Jogger in 1s
\(=\frac { 39 }{ 40 } -\frac { 34 }{ 35 } =\frac { 1365-1360 }{ 1400 } =\frac { 1 }{ 280 } m\)
Average speed of Joggers image \(=\frac { 1 }{ 280 } m/s\)
Similarly, for u = -29m, -19m and -9 m, average speed of Jogger image is \(\frac { 1 }{ 150 } m/s\),\(\frac { 1 }{ 60 } m/s\),\(=\frac { 1 }{ 10 } m/s\) respectively
The speed increases as the Jogger approaches the car>
This can be experienced by the person in the car.

Given, amplitude of an electromagnetic wave,
E₀ = 120 N/C
Frequency of wave, v = 50 MHz = 50 \(\times\) 10⁶ Hz
(i) Speed of light in vacuum, \(c=\frac{E_0}{B_0}\)
\(\begin{aligned} B_0=\frac{E_0}{c}= & \frac{120}{3 \times 10^8}=40 \times 10^{-8} \end{aligned}\)
= 400 \(\times\) 10^-9 T = 400 nT
Angular frequency of electromagnetic wave,
\(\omega=2 \pi \nu=2 \times 3.14 \times 50 \times 10^6\)
\(\omega=3.14 \times 10^8 \mathrm{rad} / \mathrm{s}\)
Wave number of electromagnetic wave,
\(k=\frac{\omega}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}\)
Wavelength of electromagnetic wave,
\(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{50 \times 10^6}=6.00 \mathrm{~m}\)
(ii) Expression of electric field, E = E₀ sin (kx - \(\omega\)t)
E = 120 sin (1.05x - 3.14 \(\times\)10⁸ t)
Expression of magnetic field B,
B = B₀ sin (kx - \(\omega\)t)
E = 120 sin (kx - \(\omega\)t)
B = 4 \(\times\)10^-7 sin (1.05x - 3.14 \(\times\) 10⁸ t)