Electromagnetic Waves MCQ Questions & Answers in Electrostatics and Magnetism | Physics
Learn Electromagnetic Waves MCQ questions & answers in Electrostatics and Magnetism are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
81.
The magnetic field of a beam emerging from a filter facing a floodlight is given by
$${B_0} = 12 \times {10^{ - 8}}\sin \left( {1.20 \times {{10}^7}z - 3.60 \times {{10}^{15}}t} \right)T.$$
What is the average intensity of the beam?
A
$$1.72 \times {10^2}\,W/{m^2}$$
B
$$1.72\,W/{m^2}$$
C
$$2.31\,W/{m^2}$$
D
$$2.31 \times {10^2}\,W/{m^2}$$
Answer :
$$1.72\,W/{m^2}$$
As we know that,
The standard equation of magnetic field
$$B = {B_0}\sin \left( {kx - \omega t} \right)$$
And, the given equation is
$$B = 12 \times {10^{ - 8}}\sin \left( {1.20 \times {{10}^7}z - 3.60 \times {{10}^{15}}t} \right)T.$$
On comparing this equation with standard equation (i), we have.
$${B_0} = 12 \times {10^{ - 8}}$$
So, the average intensity of the beam
$$\eqalign{
& {I_{{\text{av}}}} = \frac{1}{2}\frac{{B_0^2}}{{{\mu _0}}} \cdot c = \frac{1}{2} \times \frac{{{{\left( {12 \times {{10}^{ - 8}}} \right)}^2} \times 3 \times {{10}^8}}}{{4\pi \times {{10}^{ - 7}}}} \cr
& = 1.72\,W/{m^2} \cr} $$
82.
If $$c$$ is the speed of electromagnetic waves in vacuum, its speed in a medium of dielectric constant $$K$$ and relative permeability $${\mu _r}$$ is
A
$$v = \frac{1}{{\sqrt {{\mu _r}K} }}$$
B
$$v = c\sqrt {{\mu _r}K} $$
C
$$v = \frac{c}{{\sqrt {{\mu _r}K} }}$$
D
$$v = \frac{K}{{\sqrt {{\mu _r}C} }}$$
Answer :
$$v = \frac{c}{{\sqrt {{\mu _r}K} }}$$
Speed of light of vacuum $$c = \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$$ and in another medium $$v = \frac{1}{{\sqrt {\mu \varepsilon } }}$$
$$\eqalign{
& \therefore \frac{c}{v} = \sqrt {\frac{{\mu \varepsilon }}{{{\mu _0}{\varepsilon _0}}}} = \sqrt {{\mu _r}K} \cr
& \Rightarrow v = \frac{c}{{\sqrt {{\mu _r}K} }}. \cr} $$