3 Marks Question

Question 13 Marks

A point charge is moving along a straight line with a constant velocity u. Consider a small area A perpendicular to the direction of motion of the charge (El). Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large so that the electric field at any instant is essentially given by Coulomb's law.

Answer

$\text{E}=\frac{\text{Kq}}{\text{x}^2}$ [from coulomb's law]$\phi_\text{E}=\text{EA}=\frac{\text{KqA}}{\text{x}^2}$
$\text{l}_\text{d}=\in_0\frac{\text{d}\phi\text{E}}{\text{dt}}=\in_0\frac{\text{d}}{\text{dt}}\frac{\text{kqA}}{\text{x}^2}=\in_0\text{KqA}=\frac{\text{d}}{\text{dt}}\text{x}^{-2}$
$=\in_0\times\frac{1}{4\pi\in_0}\times\text{q}\times\text{A}\times-2\times\text{x}^{-3}\times\frac{\text{dx}}{\text{dt}}=\frac{\text{qAv}}{2\pi\text{x}^3}$

View full question & answer→

Question 23 Marks

A wire carries an alternating current $\text{i}=\text{i}_0\sin\omega\text{t}.$ Is there an electric field in the vicinity of the wire?

Answer

When an alternating current passes through a conductor, the changing magnetic field create a changing electric field outside it. An electromagnetic field is radiated from the surface of the conductor. There is a time-varying electric ield outside the conductor. Hence, there is a time-varying electric field in the vicinity of the wire.

View full question & answer→

Question 33 Marks

A parallel-plate capacitor of plate-area A and plate separation d is joined to a battery of emf $\in$ and internal resistance R at t = 0. Consider a plane surface of area $\frac{\text{A}}{2}$ parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.

Answer

$\text{E}=\frac{\text{Q}}{\in_0\text{A}}$ (Electric field)$\phi=\text{E.A.}=\frac{\text{Q}}{\in_0\text{A}}\frac{\text{A}}{2}=\frac{\text{Q}}{\in_02}$
$\text{i}_0=\in_0\frac{\text{d}\phi_\text{E}}{\text{dt}}=\in_0\frac{\text{d}}{\text{dt}}\Big(\frac{\text{Q}}{\in_02}\Big)$
$=\frac{1}{2}\Big(\frac{\text{dQ}}{\text{dt}}\Big)$
$=\frac{1}{2}\frac{\text{d}}{\text{dt}}\Big(\text{ECe}^{\frac{-\text{t}}{\text{RC}}}\Big)$
$=\frac{1}{2}\text{EC}-\frac{1}{\text{RC}}\text{e}^{\frac{-\text{t}}{\text{RC}}}$
$=\frac{-\text{E}}{2\text{R}}\text{e}^{\frac{-\text{td}}{\text{R}\text{E}_0\lambda}}$

View full question & answer→

Question 43 Marks

The sunlight reaching Earth has maximum electric field of $810Vm^{-1}$. What is the maximum magnetic field in this light?

Answer

$\text{E}_0=810\text{V/m},\text{ B}_0=?$We know, $\text{B}_0=\mu_0\in_0\text{CE}_0$
Putting the values,
$\text{B}_0=4\pi\times10^{-7}\times8.85\times10^{-12}\times3\times10^8\times810$
$\text{B}_0=27010.9\times10^{-10}$
$\text{B}_0=2.7\times10^{-6}\text{T}$
$\text{B}_0=2.7\mu\text{T}$

View full question & answer→

Question 53 Marks

The magnetic field in a plane electromagnetic wave is given by $\text{B}=(200\mu\text{T})\sin\Big[\big(4.0\times10^{15}\text{s}^{-1}\big)\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)\Big]$ Find the maximum electric field and the average energy density corresponding to the electric field.

Answer

$\text{B}=(200\mu\text{T})\sin\Big[\big(4.0\times10^{15}\text{s}^{-1}\big)\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)\Big]$$\text{B}_0=200\mu\text{T}$
$\text{E}_0=\text{C}\times\text{B}_0$
$\text{E}_0=200\times10^{-6}\times3\times10^{8}$
$\text{E}_0=6\times10^4$
Average energy density $=\frac{1}{2\mu_0}\text{B}_0^2=\frac{\big(200\times10^{-6}\big)^2}{2\times4\pi\times10^{-7}}$
$=\frac{4\times10^{-8}}{8\pi\times10^{-7}}=\frac{1}{20\pi}=0.0159=0.016$

View full question & answer→

Question 63 Marks

A laser beam has intensity $2.5 \times 10^{14}Wm^{-2}.$ Find amplitudes of electric and magnetic fields in the beam.

Answer

$\text{I}=2.5\times10^{14}\text{W/m}^2$We know, $\text{I}=\frac{1}{2}\in_0\text{E}^2_0\text{C}$
$\text{E}^2_0=\frac{2\text{I}}{\in_0\text{C}}$ or $\text{E}_0=\sqrt{\frac{2\text{I}}{\in_0\text{C}}}$
$\text{E}_0=\sqrt{\frac{2\times2.5\times10^{14}}{8.85\times10^{-12}\times3\times10^8}}$
$\text{E}_0=0.4339\times10^9$
$\text{E}_0=4.33\times10^8\text{N/c}$
$\text{B}_0=\mu_0\in_0\text{CE}_0$
$\text{B}_0=4\times3.14\times10^{-7}\times8.854\times10^{-12}\times3\times10^8\times4.33\times10^8$
$\text{B}_0=1.44\text{T}$

View full question & answer→

Question 73 Marks

A capacitor is connected to an alternating-current source. Is there a magnetic field between the plates?

Answer

When an alternating-current source is connected to a capacitor, the electric field between the plates of the capacitor keeps on changing with the applied voltage. Due to the changing electric field, a magnetic field exists in between the plates of the capacitor.

View full question & answer→

Question 83 Marks

Using $\text{B}=\mu_0\text{H},$ find the ratio $\frac{\text{E}_0}{\text{H}_0}$ for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.

Answer

$\text{B}=\mu_0\text{H}$$\text{H}=\frac{\text{B}}{\mu_0}$
$\frac{\text{E}_0}{\text{H}_0}=\frac{\frac{\text{B}_0}{(\mu_0\in_0\text{C})}}{\frac{\text{B}_0}{\mu_0}}=\frac{1}{\in_0\text{C}}$
$=\frac{1}{8.85\times10^{-12}\times3\times10^8}$
$=376.6\Omega=377\Omega$
$\text{Dimension}\frac{1}{\in_0\text{C}}=\frac{1}{\big[\text{LT}^{-1}\big]\big[\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^2\big]}$
$=\frac{1}{\text{M}^{-1}\text{L}^{-2}\text{T}^{3}\text{A}^2}=\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}=[\text{R}]$

View full question & answer→

Question 93 Marks

Can an electromagnetic wave be polarised?

Answer

An electromagnetic wave is a transverse wave, thus, it can be polarised. An unpolarised wave consists of many independent waves, whose planes of vibrations of electric and magnetic fields are randomly oriented. They are polarised by restricting the vibrations of the electric field vector or magnetic field vector in one direction only.

View full question & answer→

Question 103 Marks

Consider the situation of the previous problem. Define displacement resistance $\text{R}_\text{d}=\frac{\text{V}}{\text{i}_\text{d}}$ of the space between the plates, where V is the potential difference between the plates and $i_d$ is the displacement current. Show that $R_d$ varies with time as $\text{R}_\text{d}=\text{R}\big(\text{e}^{\text{t}/\tau}-1\big)$

Answer

View full question & answer→

Question 113 Marks

Can an electromagnetic wave be deflected by an electric field or a magnetic field?

Answer

No, an electromagnetic wave cannot be deflected by an electric field or a magnetic field. This is because according to Maxwell's theory, an electromagnetic wave does not interact with the static electric field and magnetic field. Even if we consider the particle nature of the wave, the photon is electrically neutral. So, it is not affected by the static magnetic and electric fields.

View full question & answer→

Physics 3 Marks Question

Take a timed test