Physics NCERT Exemplar Solutions Class 12th Chapter Eight

${\overrightarrow{B}}_{Net}=\frac{{\mu }_0}{4\pi }*2\left[\frac{i_1}{d_1}-\frac{i_2}{d_2}\right]\left(-\widehat{k}\right)$

$=10^{-7}*2\left[\frac{4}{4}*100-\frac{2}{6}*100\right]\left(-\widehat{k}\right)$

$=2*10^{-5}\left[1-\frac{1}{3}\right]-\widehat{k}$

${\overrightarrow{B}}_{Net}=\frac{4}{3}*10^{-5}\left(-\widehat{k}\right)$                        

q = 3πc

$\overrightarrow{v}=2\widehat{i}+3\widehat{j}$

$\overrightarrow{F}=q\left(\overrightarrow{v}*\overrightarrow{B}\right)$               

= $3\pi \left[\left(2\widehat{i}*3\widehat{j}\right)*\frac{4}{3}*10^{-5}\left(-\widehat{k}\right)\right]$

$\overrightarrow{F}=4\pi *10^{-5}\left(-3\widehat{i}+2\widehat{j}\right)N$  

x = 3

$F_{23}=\frac{k{q}_2{q}_3}{d^2}=\frac{k*2*2}{12}=4k$  

$F_{21}=4k$                                        

$F_{13}=4k$             

Net force on q₂ = F_R = $\sqrt[]{F_{21}^2+F_{13}^2+2F_{21}{F}_{23}cos\theta }$  

= $\sqrt[]{{\left(4k\right)}^2+{\left(4k\right)}^2+2*4kcos60°}$  

=  $k\sqrt[]{48}-\left(1\right)$

Force b/w (1) & (2) F₂₁ = 4k – (2)p

$\frac{F_R}{F_{21}}=\frac{k\sqrt[]{48}}{4k}=\frac{4\sqrt[]{3}}{4}=\sqrt[]{3}:1$

          

${?}_{flot}=\frac{q}{2{\in }_0}\left(1-cos\theta \right)$

When, 'q' is at the centre of the flat surface then, $\theta =50.$

${?}_{flat}=\frac{q}{2{\in }_0}\left(1-cos90°\right)$

$=\frac{q}{2{\in }_0}$

$x=sin\pi \left(t+\frac{1}{3}\right)$

$v=\frac{dx}{dt}=cos\left[\pi \left(t+\frac{1}{3}\right)\right]*\pi$

$=cos\left(\pi +\frac{\pi }{3}\right)*\pi$

$=-cos\frac{\pi }{3}*\left(\pi \right)$

$=-\frac{\pi }{2}=-1.57m/sec$              

Speed = 157 cm/sec

40% of K.E. = $mc\Delta T+mL$  

$0.4*\frac{1}{2}*m{v}^2=m\left[125\left(327-127\right)+2.5*10^4\right]$                

$\Rightarrow 0.2v^2=\left[250*10^2+25*10^3\right]$                

$v^2=\frac{2*25*10^3}{0.2}$                

v = 5 * 100

v = 500 m/sec

The EM wave speed in a vacuum is determined by the permeability and permittivity of free space. It is given by the formula:

$c=\frac{1}{\sqrt{?0_{}?0_{}}}$

The constant value is around $3*{10}^8\text{ m/s.}$

This speed forms the basis for many scientific and technological applications. It is a fundamental physical constant.

The electric field (E) and magnetic field (B) in an electromagnetic wave are always perpendicular to each other and to the direction of wave propagation. The EM waves are transverse in nature. When the EM waves are moving along the X-axis, then along the y-axis, the electric field may oscillate and the magnetic field along the z-axis.

Maxwell introduced the displacement current which ensures that when the conduction current is absent, the current continues in the circuits, such as in the capacitors. The displacement current plays a significant role in the derivation of electromagnetic waves. It shows that an electric field that is changing can produce a magnetic field. In free space, it enables wave propagation.

This is a multiple choice answer as classied in NCERT Exemplar

momentum per unit time per unit area = intensity/ speed of wave

 = I/c= radiation pressure (p)

Momentum is always double when a light gets reflected back as in that case the momentum which is positive to one side added to momentum which is negative to other side so momentum is always double

So it becomes 2I/c